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1.35. On the lateral sides $AB$ and $CD$ of trapezoid $ABCD$, points $M$ and $N$ are taken such that segment $MN$ is parallel to the bases and divides the area of the trapezoid in half. Find the length of $MN$, if $BC=a$ and $AD=b$. | 1.35. Let $M N=x ; E$ be the intersection point of lines $A B$ and $C D$. Triangles $E B C, E M N$ and $E A D$ are similar, so $S_{E B C}: S_{E M N}: S_{E A D}=$ $=a^{2}: x^{2}: b^{2}$. Since $S_{E M N}-S_{E B C}=S_{M B C N}=$ $=S_{M A D N}=S_{E A D}-S_{E M N}$, then $x^{2}-a^{2}=b^{2}-x^{2}$, i.e. $x^{2}=\left(a^{2}+b... | \sqrt{\frac{^2+b^2}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,678 |
1.36. Through a certain point $Q$, taken inside triangle $A B C$, three lines parallel to its sides are drawn. These lines divide the triangle into six parts, three of which are triangles with areas $S_{1}, S_{2}$, and $S_{3}$. Prove that the area of triangle $A B C$ is equal to $\left(\sqrt{S_{1}}+\sqrt{S_{2}}+\sqrt{S... | 1.36. Draw through point $Q$, taken inside triangle $A B C$, lines $D E, F G$ and $H I$ parallel to $B C, C A$ and $A B$ respectively so that points $F$ and $H$ lie on side $B C$, points $E$ and $I$ on side $A C$, and points $D$ and $G$ on side $A B$ (Fig. 1.8). Introduce the following notations: $S=S_{A B C}, S_{1}=$ ... | (\sqrt{S_{1}}+\sqrt{S_{2}}+\sqrt{S_{3}})^{2} | Geometry | proof | Yes | Yes | olympiads | false | 26,679 |
1.37. Prove that the area of a triangle with sides equal to the medians of a triangle of area $S$ is equal to $3 S / 4$. | 1.37. Let $M$ be the point of intersection of the medians of triangle $ABC$; point $A_{1}$ is symmetric to $M$ with respect to the midpoint of segment $BC$. The lengths of the sides of triangle $CMA_{1}$ are in the ratio $2:3$ to the medians of triangle $ABC$. Therefore, the desired area is $9 S_{CMA_{1}} / 4$. It is c... | 3S/4 | Geometry | proof | Yes | Yes | olympiads | false | 26,680 |
1.38. a) Prove that the area of the quadrilateral formed by the midpoints of the sides of a convex quadrilateral $ABCD$ is half the area of $ABCD$.
b) Prove that if the diagonals of a convex quadrilateral are equal, then its area is equal to the product of the lengths of the segments connecting the midpoints of the op... | 1.38. Let $E, F, G$ and $H$ be the midpoints of sides $A B, B C, C D$ and $D A$.
a) It is clear that $S_{A E H}+S_{C F G}=\frac{S_{A B D}}{4}+\frac{S_{C B D}}{4}=\frac{S_{A B C D}}{4}$. Similarly, $S_{B E F}+S_{D G H}=\frac{S_{A B C D}}{4}$. Therefore, $S_{E F G H}=S_{A B C D}-\frac{S_{A B C D}}{4}-\frac{S_{A B C D}}{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,681 |
1.39. A point $O$ inside a convex quadrilateral of area $S$ is reflected symmetrically with respect to the midpoints of its sides. Find the area of the quadrilateral with vertices at the resulting points.
## §4. Auxiliary Congruent Triangles
Translate the text above into English, please keep the original text's line ... | 1.39. Let $E, F, G$ and $H$ be the midpoints of the sides of quadrilateral $ABCD$; points $E_{1}, F_{1}, G_{1}$ and $H_{1}$ are symmetric to point $O$ with respect to these points. Since $EF$ is the midline of triangle $E_{1} O F_{1}$, then $S_{E_{1} O F_{1}}=4 S_{E O F}$. Similarly, $S_{F_{1} O G_{1}}=4 S_{F O G}, S_{... | 2S | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,682 |
1.40. The leg $BC$ of the right triangle $ABC$ with a right angle at $C$ is divided by points $D$ and $E$ into three equal parts. Prove that if $BC = 3AC$, then the sum of the angles $AEC$, $ADC$, and $ABC$ is $90^{\circ}$. | 1.40. First solution. Consider the square $B C M N$ and divide its side $M N$ into three equal parts by points $P$ and $Q$ (Fig. 1.9). Then $\triangle A B C = \triangle P D Q$ and $\triangle A C D = \triangle P M A$. Therefore, triangle $P A D$ is an isosceles right triangle and $\angle A B C + \angle A D C = \angle P ... | 90 | Geometry | proof | Yes | Yes | olympiads | false | 26,683 |
1.41. Point $K$ is the midpoint of side $A B$ of square $A B C D$, and point $L$ divides diagonal $A C$ in the ratio $A L: L C=3: 1$. Prove that angle $K L D$ is a right angle. | 1.41. Drop perpendiculars $L M$ from point $L$ to $A B$ and $L N$ to $A D$. Then $K M=M B=N D$ and $K L=L B=D L$, so the right triangles $K M L$ and $D N L$ are equal. Therefore, $\angle D L K=\angle N L M=90^{\circ}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,684 |
1.42. Through the vertex $A$ of the square $A B C D$, lines $l_{1}$ and $l_{2}$ are drawn, intersecting its sides. Perpendiculars $B B_{1}$, $B B_{2}$, $D D_{1}$, and $D D_{2}$ are dropped from points $B$ and $D$ to these lines. Prove that the segments $B_{1} B_{2}$ and $D_{1} D_{2}$ are equal and perpendicular. | 1.42. Since $D_{1} A=B_{1} B, A D_{2}=B B_{2}$ and $\angle D_{1} A D_{2}=\angle B_{1} B B_{2}$, then $\triangle D_{1} A D_{2}=$ $=\triangle B_{1} B B_{2}$. The sides $A D_{1}$ and $B B_{1}$ (as well as $A D_{2}$ and $B B_{2}$) of these triangles are perpendicular, so $B_{1} B_{2} \perp D_{1} D_{2}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,685 |
1.43. On the legs $CA$ and $CB$ of an isosceles right triangle $ABC$, points $D$ and $E$ are chosen such that $CD = CE$. The extensions of the perpendiculars dropped from points $D$ and $C$ to the line $AE$ intersect the hypotenuse $AB$ at points $K$ and $L$. Prove that $KL = LB$. | 1.43. On the extension of segment $A C$ beyond point $C$, take point $M$ such that $C M = C E$ (Fig. 1.10). Then triangle $A C E$ transforms into triangle $B C M$ when rotated $90^{\circ}$ around point $C$. Therefore, line $M B$ is perpendicular to line $A E$, and thus parallel to line $C L$. Since $M C = C E = D C$ an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,686 |
1.44*. On the sides $AB, BC, CD$, and $DA$ of the inscribed quadrilateral $ABCD$, whose lengths are $a, b, c$, and $d$, rectangles of sizes $a \times c, b \times d, c \times a$, and $d \times b$ are constructed externally. Prove that their centers are the vertices of a rectangle. | 1.44. Let rectangles $A B C_{1} D_{1}$ and $A_{2} B C D_{2}$ be constructed on sides $A B$ and $B C$; $P, Q, R$ and $S$ are the centers of the rectangles constructed on sides $A B, B C, C D$ and $D A$. Since $\angle A B C + \angle A D C = 180^{\circ}$, then $\triangle A D C = \triangle A_{2} B C_{1}$, and thus $\triang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,687 |
1.45*. Hexagon $A B C D E F$ is inscribed in a circle of radius $R$ with center $O$, and $A B=C D=E F=R$. Prove that the points of pairwise intersection of the circumcircles of triangles $B O C, D O E$, and $F O A$, distinct from point $O$, are the vertices of an equilateral triangle with side $R$.
$$
* * *
$$ | 1.45. Let $K, L, M$ be the points of intersection of the circumcircles of triangles $F O A$ and $B O C, B O C$ and $D O E, D O E$ and $F O A$; $2 \alpha, 2 \beta$ and $2 \gamma$ are the angles at the vertices of the isosceles triangles $B O C, D O E$ and $F O A$ (Fig. 1.11). The point $K$ lies on the arc $O B$ of the c... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,688 |
1.46. On the sides $B C$ and $C D$ of the parallelogram $A B C D$, regular triangles $B C K$ and $D C L$ are constructed externally. Prove that triangle $A K L$ is regular. | 1.46. Let $\angle A=\alpha$. It is easy to verify that both angles $K C L$ and $A D L$ are equal to $240^{\circ}-\alpha$ (or $120^{\circ}+\alpha$). And since $K C=B C=A D$ and $C L \neq D L$, then $\triangle K C L = \triangle A D L$, which means $K L=A L$. Similarly, $K L=A K$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,689 |
1.47. On the sides of a parallelogram, squares are constructed externally. Prove that their centers form a square. | 1.47. Let $P, Q$ and $R$ be the centers of squares constructed on the sides $D A, A B$ and $B C$ of a parallelogram with an acute angle $\alpha$ at vertex $A$. It is easy to verify that $\angle P A Q=90^{\circ}+\alpha=\angle R B Q$, and therefore, $\triangle P A Q=\triangle R B Q$. The sides $A Q$ and $B Q$ of these tr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,690 |
1.48*. On the sides of an arbitrary triangle $ABC$, isosceles triangles with angles $2 \alpha, 2 \beta$, and $2 \gamma$ at vertices $A^{\prime}, B^{\prime}$, and $C^{\prime}$ are constructed externally, and $\alpha+\beta+\gamma=180^{\circ}$. Prove that the angles of triangle $A^{\prime} B^{\prime} C^{\prime}$ are $\alp... | 1.48. First, note that the sum of the angles at vertices $A, B$, and $C$ of the hexagon $A B^{\prime} C A^{\prime} B C^{\prime}$ is $360^{\circ}$, since the sum of its angles at the other vertices is $360^{\circ}$ by the problem's condition. Construct an external triangle $A C^{\prime} P$ on side $A C^{\prime}$, equal ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,691 |
1.49*. On the sides of triangle $A B C$, similar isosceles triangles $A B_{1} C$ and $A C_{1} B$ are constructed externally, and $B A_{1} C$ is constructed internally. Prove that $A B_{1} A_{1} C_{1}$ is a parallelogram. | 1.49. Since $B A: B C=B C_{1}: B A_{1}$ and $\angle A B C=\angle C_{1} B A_{1}$, then $\triangle A B C \sim \triangle C_{1} B A_{1}$. Similarly, $\triangle A B C \sim \triangle B_{1} A_{1} C$. And since $B A_{1}=A_{1} C$, then
 On the sides $AB$ and $AC$ of triangle $ABC$, right triangles $ABC_1$ and $AB_1C$ are constructed externally, such that $\angle C_1 = \angle B_1 = 90^\circ$, $\angle ABC_1 = \angle ACB_1 = \varphi$; $M$ is the midpoint of $BC$. Prove that $MB_1 = MC_1$ and $\angle B_1MC_1 = 2\varphi$.
b) On the sides of tria... | 1.50. a) Let $P$ and $Q$ be the midpoints of sides $AB$ and $AC$. Then $MP = AC / 2 = QB_1$, $MQ = AB / 2 = PC_1$, and $\angle C_1PM = \angle C_1PB + \angle BPM = \angle B_1QC + \angle CQM = \angle B_1QM$. Therefore, $\triangle MQB_1 = \triangle C_1PM$, which means $MC_1 = MB_1$. Moreover, $\angle PM C_1 + \angle QM B_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,693 |
1.51*. On the unequal sides $AB$ and $AC$ of triangle $ABC$, isosceles triangles $AC_1B$ and $AB_1C$ are constructed externally with an angle $\varphi$ at the vertex.
a) $M$ is a point on the median $AA_1$ (or its extension), equidistant from points $B_1$ and $C_1$. Prove that $\angle B_1MC_1 = \varphi$.
b) $O$ is a p... | 1.51. a) Let $B'$ be the intersection point of line $AC$ and the perpendicular to line $AB_1$ drawn from point $B_1$; point $C'$ is defined similarly. Since $AB' : AC' = AC_1 : AB_1 = AB : AC$, it follows that $B'C' \| BC$. If $N$ is the midpoint of segment $B'C'$, then, as follows from problem 1.51, $NC_1 = NB_1$ (i.e... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,694 |
1.52*. On the sides of a convex quadrilateral $ABCD$, similar rhombi are constructed externally, with their acute angles $\alpha$ adjacent to vertices $A$ and $C$. Prove that the segments connecting the centers of opposite rhombi are equal, and the angle between them is $\alpha$.
## §5. Triangle Formed by the Bases of... | 1.52. Let $O_{1}, O_{2}, O_{3}$ and $O_{4}$ be the centers of rhombuses constructed on the sides $A B, B C, C D$ and $D A$; $M$ is the midpoint of the diagonal $A C$. Then $M O_{1}=M O_{2}$ and $\angle O_{1} M O_{2}=\alpha$ (see problem 1.51). Similarly, $M O_{3}=M O_{4}$ and $\angle O_{3} M O_{4}=\alpha$. Therefore, w... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,695 |
1.53. Let $A A_{1}$ and $B B_{1}$ be the altitudes of triangle $A B C$. Prove that $\triangle A_{1} B_{1} C \sim \triangle A B C$. What is the similarity ratio? | 1.53. Since $A_{1} C=A C|\cos C|, B_{1} C=B C|\cos C|$ and angle $C$ is common to triangles $A B C$ and $A_{1} B_{1} C$, these triangles are similar, with the similarity coefficient equal to $|\cos C|$. | |\cosC| | Geometry | proof | Yes | Yes | olympiads | false | 26,696 |
1.54. From the vertex $C$ of an acute-angled triangle $ABC$, a height $CH$ is dropped, and from the point $H$, perpendiculars $HM$ and $HN$ are dropped to the sides $BC$ and $AC$ respectively. Prove that $\triangle MNC \sim \triangle ABC$. | 1.54. Since points $M$ and $N$ lie on the circle with diameter $C H$, then $\angle C M N = \angle C H N$, and since $A C \perp H N$, then $\angle C H N = \angle A$. Similarly, $\angle C N M = \angle B$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,697 |
1.55. In triangle $ABC$, altitudes $BB_1$ and $CC_1$ are drawn.
a) Prove that the tangent at point $A$ to the circumcircle is parallel to the line $B_1 C_1$.
b) Prove that $B_1 C_1 \perp OA$, where $O$ is the center of the circumcircle. | 1.55. a) Let $l$ be the tangent at point $A$ to the circumcircle. Then $\angle(l, A B)=\angle(A C, C B)=\angle\left(C_{1} B_{1}, A C_{1}\right)$, which means $l \| B_{1} C_{1}$.
b) Since $O A \perp l$ and $l \| B_{1} C_{1}$, then $O A \perp B_{1} C_{1}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,698 |
1.56. On the sides of an acute-angled triangle $A B C$, points $A_{1}$, $B_{1}$, and $C_{1}$ are taken such that the segments $A A_{1}$, $B B_{1}$, and $C C_{1}$ intersect at point $H$. Prove that $A H \cdot A_{1} H = B H \cdot B_{1} H = C H \cdot C_{1} H$ if and only if $H$ is the orthocenter of triangle $A B C$. | 1.56. If $A A_{1}, B B_{1}$ and $C C_{1}$ are altitudes, then the right triangles $A A_{1} C$ and $B B_{1} C$ have equal angles at vertex $C$, so they are similar. Therefore, $\triangle A_{1} B H \sim \triangle B_{1} A H$, which means $A H \cdot A_{1} H = B H \cdot B_{1} H$. Similarly, $B H \cdot B_{1} H = C H \cdot C_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,699 |
1.57. a) Prove that the altitudes $A A_{1}, B B_{1}$ and $C C_{1}$ of an acute-angled triangle $A B C$ bisect the angles of triangle $A_{1} B_{1} C_{1}$.
b) On the sides $A B, B C$ and $C A$ of an acute-angled triangle $A B C$, points $C_{1}, A_{1}$ and $B_{1}$ are taken respectively. Prove that if $\angle B_{1} A_{1} ... | 1.57. a) According to problem $1.54 \angle C_{1} A_{1} B=\angle C A_{1} B_{1}=\angle A$. Since $A A_{1} \perp B C$, then $\angle C_{1} A_{1} A=\angle B_{1} A_{1} A$. Similarly, it can be proved that the rays $B_{1} B$ and $C_{1} C$ are the bisectors of the angles $A_{1} B_{1} C_{1}$ and $A_{1} C_{1} B_{1}$.
b) The lin... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,700 |
1.59. In an acute-angled triangle $A B C$, the altitudes $A A_{1}, B B_{1}$, and $C C_{1}$ are drawn. Prove that if $A_{1} B_{1} \| A B$ and $B_{1} C_{1} \| B C$, then $A_{1} C_{1} \| A C$. | 1.59. According to problem $1.54 \angle B_{1} A_{1} C=\angle B A C$. Since $A_{1} B_{1} \| A B$, then $\angle B_{1} A_{1} C=\angle A B C$. Therefore, $\angle B A C=\angle A B C$. Similarly, from $B_{1} C_{1} \| B C$, it follows that $\angle A B C=\angle B C A$. Therefore, triangle $A B C$ is equilateral and $A_{1} C_{1... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,702 |
1.60*. Let $p$ be the semiperimeter of an acute-angled triangle $ABC$, and $q$ be the semiperimeter of the triangle formed by the bases of its altitudes. Prove that $p: q = R: r$, where $R$ and $r$ are the radii of the circumcircle and incircle of triangle $ABC$.
## §6. Similar Figures | 1.60. Let $O$ be the center of the circumcircle of triangle $ABC$. Since $OA \perp B_{1}C_{1}$ (see problem 1.56, b), then $S_{AOC_{1}} + S_{AOB_{1}} = R \cdot B_{1}C_{1} / 2$. Similar reasoning for vertices $B$ and $C$ shows that $S_{ABC} = qR$. On the other hand, $S_{ABC} = pr$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,703 |
1.61. A circle of radius $r$ is inscribed in a triangle. Tangents to this circle, parallel to the sides of the triangle, cut off three smaller triangles from it. Let $r_{1}, r_{2}, r_{3}$ be the radii of the circles inscribed in these triangles. Prove that $r_{1}+r_{2}+r_{3}=r$. | 1.61. The perimeter of the triangle cut off by a line parallel to side $BC$ is equal to the sum of the distances from point $A$ to the points of tangency of the inscribed circle with sides $AB$ and $AC$, so the sum of the perimeters of the small triangles is equal to the perimeter of triangle $ABC: P_{1}+P_{2}+P_{3}=P$... | r_{1}+r_{2}+r_{3}=r | Geometry | proof | Yes | Yes | olympiads | false | 26,704 |
1.62. Given a triangle $A B C$. Construct two lines $x$ and $y$ such that for any point $M$ on side $A C$, the sum of the lengths of segments $M X_{M}$ and $M Y_{M}$, drawn from point $M$ parallel to lines $x$ and $y$ until they intersect sides $A B$ and $B C$ of the triangle, equals 1. | 1.62. Let \( M = A \). Then \( X_{A} = A \), so \( A Y_{A} = 1 \). Similarly, \( C X_{C} = 1 \). We will prove that \( y = A Y_{A} \) and \( x = C X_{C} \) are the desired lines. Take a point \( D \) on side \( B C \) such that \( A B \parallel M D \) (Fig. 1.13). Let \( E \) be the intersection point of the lines \( C... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,705 |
1.63. In an isosceles triangle $ABC$, from the midpoint $H$ of the base $BC$, a perpendicular $HE$ is dropped to the lateral side $AC$; $O$ is the midpoint of the segment $HE$. Prove that the lines $AO$ and $BE$ are perpendicular. | 1.63. Let $D$ be the midpoint of segment $B H$. Since $\triangle B H A \sim \triangle H E A$, then $A D: A O=A B: A H$ and $\angle D A H=\angle O A E$. Therefore, $\angle D A O=\angle B A H$, which means $\triangle D A O \sim \triangle B A H$ and $\angle D O A=\angle B A H=90^{\circ}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,706 |
1.64. Prove that the projections of the foot of a triangle's altitude onto the sides containing it and onto the other two altitudes lie on the same line. | 1.64. Let $A A_{1}, B B_{1}$, and $C C_{1}$ be the altitudes of triangle $A B C$. Drop perpendiculars $B_{1} K$ and $B_{1} N$ from point $B_{1}$ to sides $A B$ and $B C$, and perpendiculars $B_{1} L$ and $B_{1} M$ to altitudes $A A_{1}$ and $C C_{1}$. Since $K B_{1}: C C_{1}=A B_{1}: A C=L B_{1}: A_{1} C$, it follows t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,707 |
1.65. On the segment $A C$, a point $B$ is taken, and on the segments $A B, B C, C A$, semicircles $S_{1}, S_{2}, S_{3}$ are constructed on the same side of $A C$. $D$ is a point on $S_{3}$ such that $B D \perp A C$. The common tangent to $S_{1}$ and $S_{2}$ touches these semicircles at points $F$ and $E$ respectively.... | 1.65. a) Let $O$ be the midpoint of $AC$, $O_{1}$ the midpoint of $AB$, and $O_{2}$ the midpoint of $BC$. We will assume that $AB \leqslant BC$. Draw a line $O_{1}K$ through point $O_{1}$ parallel to $EF$ (where $K$ is a point on segment $EO_{2}$). We will prove that the right triangles $DBO$ and $O_{1}KO_{2}$ are equa... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,708 |
1.66*. From an arbitrary point $M$ on the circle circumscribed around the rectangle $A B C D$, perpendiculars $M Q$ and $M P$ are dropped to its two opposite sides, and perpendiculars $M R$ and $M T$ are dropped to the extensions of the other two sides. Prove that the lines $P R$ and $Q T$ are perpendicular, and the po... | 1.66. Let $M Q$ and $M P$ be the perpendiculars dropped to the sides $A D$ and $B C$, and $M R$ and $M T$ be the perpendiculars dropped to the extensions of the sides $A B$ and $C D$ (Fig. 1.14). Denote by $M_{1}$ and $P_{1}$ the second points of intersection of the lines $R T$ and $Q P$ with the circle.
Since $T M_{1... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,709 |
1.67*. To two circles, one located outside the other, one external and one internal tangent are drawn. Consider two lines, each passing through the points of tangency belonging to one of the circles. Prove that the point of intersection of these lines lies on the line connecting the centers of the circles.
## Problems... | 1.67. Let the centers of the circles be denoted as $O_{1}$ and $O_{2}$. The external tangent touches the first circle at point $K$ and the second circle at point $L$; the internal tangent touches the first circle at point $M$ and the second circle at point $N$ (Fig. 1.15). Let the lines $K M$ and $L N$ intersect the li... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,710 |
2.1. Vertex $A$ of an acute-angled triangle $ABC$ is connected to the center $O$ of the circumscribed circle by a segment. From vertex $A$, the altitude $AH$ is drawn. Prove that $\angle BAH = \angle OAC$. | 2.1. Let's draw the diameter $A D$. Then $\angle C D A = \angle C B A$, and therefore, $\angle B A H = \angle D A C$, since $\angle B H A = \angle A C D = 90^{\circ}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,711 |
2.2. Two circles intersect at points $M$ and $K$. Through $M$ and $K$, lines $A B$ and $C D$ are drawn, intersecting the first circle at points $A$ and $C$, and the second circle at points $B$ and $D$. Prove that $A C \| B D$. | 2.2. From the properties of directed angles we have $\angle(A C, C K)=\angle(A M, M K)=$ $=\angle(B M, M K)=\angle(B D, D K)=\angle(B D, C K)$, i.e., $A C \| B D$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,712 |
2.3. From an arbitrary point $M$, lying inside a given angle with vertex $A$, perpendiculars $M P$ and $M Q$ are dropped to the sides of the angle. From point $A$, a perpendicular $A K$ is dropped to the segment $P Q$. Prove that $\angle P A K = \angle M A Q$. | 2.3. Points $P$ and $Q$ lie on the circle with diameter $A M$. Therefore, $\angle Q M A = \angle Q P A$ as angles subtended by the same arc. Triangles $P A K$ and $M A Q$ are right triangles, hence $\angle P A K = \angle M A Q$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,713 |
2.4. a) The continuation of the bisector of angle $B$ of triangle $A B C$ intersects the circumscribed circle at point $M$; $O$ is the center of the inscribed circle, and $O_{b}$ is the center of the excircle touching side $A C$. Prove that points $A, C, O$, and $O_{b}$ lie on a circle with center $M$.
b) Point $O$, l... | 2.4. a) Since $\angle A O M=\angle B A O+\angle A B O=(\angle A+\angle B) / 2$ and $\angle O A M=\angle O A C + \angle C A M=\angle A / 2+\angle C B M=(\angle A+\angle B) / 2$, then $M A=M O$. Similarly, $M C=M O$.
Since triangle $O A O_{b}$ is a right triangle and $\angle A O M=\angle M A O=\varphi$, then $\angle M A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,714 |
2.5. Vertices $A$ and $B$ of triangle $A B C$ with right angle $C$ slide along the sides of a right angle with vertex $P$. Prove that point $C$ moves along a segment. | 2.5. Points $P$ and $C$ lie on the circle with diameter $A B$, so $\angle A P C = \angle A B C$, i.e., the measure of angle $A P C$ is constant.
Remark. A similar statement is true for any triangle $A B C$, the vertices of which slide along the sides of angle $M P N$, equal to $180^{\circ} - \angle C$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,715 |
2.6. The diagonal $AC$ of the square $ABCD$ coincides with the hypotenuse of the right triangle $ACK$, and points $B$ and $K$ lie on the same side of the line $AC$. Prove that $BK=|AK-CK| / \sqrt{2}$ and $DK=(AK+CK) / \sqrt{2}$. | 2.6. Points $B, D$ and $K$ lie on a circle with diameter $A C$. Let, for definiteness, $\angle K C A=\varphi \leqslant 45^{\circ}$. Then $B K=A C \sin \left(45^{\circ}-\varphi\right)=A C(\cos \varphi-\sin \varphi) / \sqrt{2}$ and $D K=A C \sin \left(45^{\circ}+\varphi\right)=A C(\cos \varphi+\sin \varphi) / \sqrt{2}$. ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,716 |
2.7. In triangle $A B C$, medians $A A_{1}$ and $B B_{1}$ are drawn. Prove that if $\angle C A A_{1}=\angle C B B_{1}$, then $A C=B C$. | 2.7. Since $\angle B_{1} A A_{1}=\angle A_{1} B B_{1}$, points $A, B, A_{1}$ and $B_{1}$ lie on the same circle. Parallel lines $A B$ and $A_{1} B_{1}$ intercept equal chords $A B_{1}$ and $B A_{1}$ on it. Therefore, $A C=B C$. | AC=BC | Geometry | proof | Yes | Yes | olympiads | false | 26,717 |
2.8. All angles of triangle $ABC$ are less than $120^{\circ}$. Prove that there exists a point inside it from which all sides of the triangle are seen at an angle of $120^{\circ}$.
The point from problem 2.8 is called the Torricelli point. | 2.8. Construct an equilateral triangle \(A_{1} B C\) externally on the side \(B C\) of triangle \(A B C\). Let \(P\) be the point of intersection of the line \(A A_{1}\) with the circumcircle of triangle \(A_{1} B C\). Then the point \(P\) lies inside triangle \(A B C\) and \(\angle A P C=180^{\circ}-\angle A_{1} P C=1... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,718 |
2.9. A circle is divided into equal arcs by $n$ diameters. Prove that the bases of the perpendiculars dropped from an arbitrary point $M$ inside the circle to these diameters are the vertices of a regular polygon. | 2.9. The feet of the perpendiculars dropped from point $M$ to the diameters lie on the circle $S$ with diameter $O M$ (where $O$ is the center of the original circle). The intersection points of these diameters with circle $S$, different from point $O$, divide it into $n$ arcs. Since all arcs not containing point $O$ s... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,719 |
2.10. On a circle, points $A, B, M$ and $N$ are given. From point $M$, chords $M A_{1}$ and $M B_{1}$ are drawn, perpendicular to the lines $N B$ and $N A$ respectively. Prove that $A A_{1} \| B B_{1}$. | 2.10. It is clear that $\angle\left(A A_{1}, B B_{1}\right)=\angle\left(A A_{1}, A B_{1}\right)+\angle\left(A B_{1}, B B_{1}\right)=\angle\left(M A_{1}, M B_{1}\right)+$ $+\angle(A N, B N)$. Since $M A_{1} \perp B N$ and $M B_{1} \perp A N$, then $\angle\left(M A_{1}, M B_{1}\right)=\angle(B N, A N)=$ $=-\angle(A N, B ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,720 |
2.11. A hexagon $A B C D E F$ is inscribed, and $A B \| D E$ and $B C \| E F$. Prove that $C D \| A F$. | 2.11. Since $A B \| D E$, then $\angle A C E=\angle B F D$, and since $B C \| E F$, then $\angle C A E=$ $=\angle F D B$. Triangles $A C E$ and $B D F$ have two equal angles each, so their third angles are also equal. From the equality of these angles, it follows that the arcs $A C$ and $D F$ are equal, i.e., the chord... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,721 |
2.12*. The polygon $A_{1} A_{2} \ldots A_{2 n}$ is inscribed. It is known that all pairs of its opposite sides, except one, are parallel. Prove that if $n$ is odd, the remaining pair of sides is also parallel, and if $n$ is even, the remaining pair of sides is equal in length. | 2.12. We will prove the statement by induction on $n$. For a quadrilateral, the statement is obvious, and for a hexagon, it was proven in the previous problem. Assume that the statement is proven for a $2(n-1)$-gon, and we will prove it for a $2n$-gon. Let $A_{1} \ldots A_{2n}$ be a $2n$-gon, in which $A_{1} A_{2} \| A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,722 |
2.13*. Given a triangle $A B C$. Prove that there exist two families of equilateral triangles, the sides of which (or their extensions) pass through points $A, B$ and $C$. Also prove that the centers of the triangles in these families lie on two concentric circles.
## §2. The Measure of the Angle Between Two Chords
S... | 2.13. Let the lines $F G, G E$ and $E F$ pass through points $A, B$ and $C$, and the triangle $E F G$ is equilateral, i.e., $\angle(G E, E F)=\angle(E F, F G)=\angle(F G, G E) = \pm 60^{\circ}$. Then $\angle(B E, E C)=\angle(C F, F A)=\angle(A G, G B) = \pm 60^{\circ}$. Choosing one of the signs, we obtain three circle... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,723 |
2.14. On a circle, points $A, B, C, D$ are given in the specified order. $M$ is the midpoint of arc $AB$. Denote the points of intersection of chords $MC$ and $MD$ with chord $AB$ as $E$ and $K$. Prove that $K E C D$ is a cyclic quadrilateral. | 2.14. It is clear that $2(\angle K E C+\angle K D C)=(\smile M B+\smile A C)+(\smile M B+\smile B C)=360^{\circ}$, since $\smile M B=\smile A M)$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,724 |
2.15. A circle with a radius equal to the height of an equilateral triangle rolls along one of its sides. Prove that the angular measure of the arc cut off on the circle by the sides of the triangle is always $60^{\circ}$. | 2.15. Let the angular measure of the arc cut off by the sides of triangle $ABC$ on the circle be denoted by $\alpha$. Consider the arc cut off by the extensions of the sides of the triangle on the circle, and denote its angular measure by $\alpha^{\prime}$. Then $\left(\alpha+\alpha^{\prime}\right) / 2=\angle B A C=60^... | 60 | Geometry | proof | Yes | Yes | olympiads | false | 26,725 |
2.16. The diagonals of an isosceles trapezoid $A B C D$ with lateral side $A B$ intersect at point $P$. Prove that the center $O$ of its circumscribed circle lies on the circumscribed circle of triangle $A P B$.
Translate the above text into English, please retain the original text's line breaks and format, and output... | 2.16. Since $\angle A P B=(\smile A B+\smile C D) / 2=\angle A O B$, point $O$ lies on the circumcircle of triangle $A P B$. | Number Theory | proof | Yes | Yes | olympiads | false | 26,726 | |
2.17. On a circle, points $A, B, C, D$ are given in the specified order; $A_{1}, B_{1}, C_{1}$, and $D_{1}$ are the midpoints of the arcs $A B, B C, C D$, and $D A$, respectively. Prove that $A_{1} C_{1} \perp B_{1} D_{1$. | 2.17. Let $O$ be the point of intersection of the lines $A_{1} C_{1}$ and $B_{1} D_{1} ; \alpha, \beta, \gamma$ and $\delta$ be the angular measures of the arcs $A B, B C, C D$ and $D A$. Then $\angle A_{1} O B_{1}=\left(\smile A_{1} B+\smile B B_{1}+\smile C_{1} D+\right.$ $\left.+\smile D D_{1}\right) / 2=(\alpha+\be... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,727 |
2.18. Inside triangle $A B C$, a point $P$ is taken such that $\angle B P C = \angle A + 60^{\circ}, \angle A P C = \angle B + 60^{\circ}$, and $\angle A P B = \angle C + 60^{\circ}$. The lines $A P, B P$, and $C P$ intersect the circumcircle of triangle $A B C$ at points $A^{\prime}, B^{\prime}$, and $C^{\prime}$. Pro... | 2.18. Adding the equalities $\smile C^{\prime} A+\smile C A^{\prime}=2\left(180^{\circ}-\angle A P C\right)=240^{\circ}-2 \angle B$ and $\smile A B^{\prime}+\smile B A^{\prime}=240^{\circ}-2 \angle C$, and then subtracting from their sum the equality $\smile B A^{\prime}+$ $+\smile C A^{\prime}=2 \angle A$, we get $\sm... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,728 |
2.19. Points $A, C_{1}, B, A_{1}, C, B_{1}$ are taken on a circle in the given order.
a) Prove that if the lines $A A_{1}, B B_{1}$, and $C C_{1}$ are the angle bisectors of triangle $A B C$, then they are the altitudes of triangle $A_{1} B_{1} C_{1}$.
b) Prove that if the lines $A A_{1}, B B_{1}$, and $C C_{1}$ are ... | 2.19. a) For example, let's prove that $A A_{1} \perp C_{1} B_{1}$. Let $M$ be the point of intersection of these segments. Then $\angle A M B_{1}=\left(\smile A B_{1}+\smile A_{1} B+\smile B C_{1}\right) / 2=\angle A B B_{1}+\angle A_{1} A B+$ $+\angle B C C_{1}=(\angle B+\angle A+\angle C) / 2=90^{\circ}$.
b) Let $M... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,729 |
2.20. Triangles \(T_{1}\) and \(T_{2}\) are inscribed in a circle, and the vertices of triangle \(T_{2}\) are the midpoints of the arcs into which the circle is divided by the vertices of triangle \(T_{1}\). Prove that in the hexagon formed by the intersection of triangles \(T_{1}\) and \(T_{2}\), the diagonals connect... | 2.20. Let the vertices of triangle $T_{1}$ be $A, B$, and $C$; the midpoints of the arcs $BC, CA, AB$ be $A_{1}, B_{1}, C_{1}$. Then $T_{2}=A_{1} B_{1} C_{1}$. The lines $A A_{1}, B B_{1}, C C_{1}$ are the angle bisectors of triangle $T_{1}$, so they intersect at a single point $O$. Let the lines $AB$ and $C_{1} B_{1}$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,730 |
2.21. Two circles intersect at points $P$ and $Q$. Through point $A$ of the first circle, lines $A P$ and $A Q$ are drawn, intersecting the second circle at points $B$ and $C$. Prove that the tangent at point $A$ to the first circle is parallel to the line $B C$. | 2.21. Let $l$ be the tangent at point $A$ to the first circle. Then $\angle(l, A P)=$ $=\angle(A Q, P Q)=\angle(B C, P B)$, which means $l \| B C$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,731 |
2.22. Circles $S_{1}$ and $S_{2}$ intersect at points $A$ and $P$. Through point $A$, a tangent $A B$ is drawn to circle $S_{1}$, and through point $P$, a line $C D$ is drawn parallel to $A B$ (points $B$ and $C$ lie on $S_{2}$, point $D$ lies on $S_{1}$). Prove that $A B C D$ is a parallelogram. | 2.22. Since $\angle(A B, A D)=\angle(A P, P D)=\angle(A B, B C)$, then $B C \| A D$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,732 |
2.23. Circles $S_{1}$ and $S_{2}$ intersect at points $A$ and $B$. Through point $A$, a tangent $A Q$ is drawn to circle $S_{1}$ (point $Q$ lies on $S_{2}$), and through point $B$, a tangent $B S$ is drawn to circle $S_{2}$ (point $S$ lies on $S_{1}$). Lines $B Q$ and $A S$ intersect circles $S_{1}$ and $S_{2}$ at poin... | 2.23. From the equality of the angle between the tangent and the chord to the corresponding inscribed angle, it follows that $\angle(A B, B C)=\angle(A Q, Q B)$ and $\angle(B A, A Q)=\angle(B S, S A)$. Therefore, $\angle(B A, A C)=\angle(A B, B Q)$, which means $P S \| Q R$. Further, $\angle(A P, P Q)=$ $=\angle(A B, B... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,733 |
2.24. The tangent at point $A$ to the circumcircle of triangle $A B C$ intersects line $B C$ at point $E$; $A D$ is the angle bisector of triangle $A B C$. Prove that $A E=E D$. | 2.24. Let point $E$ lie on ray $B C$ for definiteness. Then $\angle A B C = \angle E A C$ and $\angle A D E = \angle A B C + \angle B A D = \angle E A C + \angle C A D = \angle D A E$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,734 |
2.25. Circles $S_{1}$ and $S_{2}$ intersect at point $A$. Through point $A$, a line is drawn intersecting $S_{1}$ at point $B$ and $S_{2}$ at point $C$. Tangents to the circles are drawn at points $C$ and $B$, intersecting at point $D$. Prove that the angle $B D C$ does not depend on the choice of the line passing thro... | 2.25. Let $P$ be the second intersection point of the circles. Then $\angle(A B, D B) = \angle(P A, P B)$ and $\angle(D C, A C) = \angle(P C, P A)$. Adding these equalities, we get $\angle(D C, D B) = \angle(P C, P B) = \angle(P C, C A) + \angle(B A, P B)$; the last two angles subtend constant arcs. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,735 |
2.26. Two circles intersect at points $A$ and $B$. From point $A$, tangents $A M$ and $A N$ are drawn to these circles ($M$ and $N$ are points on the circles). Prove that:
a) $\angle A B N + \angle M A N = 180^{\circ}$
b) $B M / B N = (A M / A N)^{2}$. | 2.26. a) Since $\angle M A B = \angle B N A$, the sum of angles $A B N$ and $M A N$ is equal to the sum of the angles of triangle $A B N$.
b) Since $\angle B A M = \angle B N A$ and $\angle B A N = \angle B M A$, then $\triangle A M B \sim \triangle N A B$, and therefore, $A M: N A = M B: A B$ and $A M: N A = A B: N B... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,736 |
2.27. Two circles touch internally at point $M$. Let $A B$ be a chord of the larger circle that touches the smaller circle at point $T$. Prove that $M T$ is the bisector of angle $A M B$. | 2.27. Let $A_{1}$ and $B_{1}$ be the points of intersection of the lines $M A$ and $M B$ with the smaller circle. Since $M$ is the center of homothety of the circles, $A_{1} B_{1} \| A B$. Therefore, $\angle A_{1} M T=\angle A_{1} T A=\angle B_{1} A_{1} T=\angle B_{1} M T$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,737 |
2.28. Through a point $M$ inside a circle $S$, a chord $A B$ is drawn; from the point $M$, perpendiculars $M P$ and $M Q$ are dropped to the tangents passing through points $A$ and $B$. Prove that the value $1 / P M + 1 / Q M$ does not depend on the choice of the chord passing through the point $M$. | 2.28. Let $\varphi$ be the angle between the chord $A B$ and the tangent passing through one of its ends. Then $A B=2 R \sin \varphi$, where $R$ is the radius of the circle $S$. Moreover, $P M=A M \sin \varphi$ and $Q M=B M \sin \varphi$. Therefore, $\frac{1}{P M}+\frac{1}{Q M}=((A M+$ $+B M) / \sin \varphi) A M \cdot ... | \frac{2R}{AM\cdotBM} | Geometry | proof | Yes | Yes | olympiads | false | 26,738 |
2.29. Circle $S_{1}$ touches the sides of angle $A B C$ at points $A$ and $C$. Circle $S_{2}$ touches the line $A C$ at point $C$ and passes through point $B$, intersecting circle $S_{1}$ at point $M$. Prove that line $A M$ bisects segment $B C$. | 2.29. Let the line $A M$ intersect the circle $S_{2}$ at point $D$. Then $\angle M D C = \angle M C A = \angle M A B$, so $C D \| A B$. Furthermore, $\angle C A M = \angle M C B = \angle M D B$, so $A C \| B D$. Therefore, $A B C D$ is a parallelogram, and its diagonal $A D$ bisects the diagonal $B C$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,739 |
2.30. Circle $S$ touches circles $S_{1}$ and $S_{2}$ at points $A_{1}$ and $A_{2}$; $B$ is a point on circle $S$, and $K_{1}$ and $K_{2}$ are the second points of intersection of lines $A_{1} B$ and $A_{2} B$ with circles $S_{1}$ and $S_{2}$. Prove that if line $K_{1} K_{2}$ is tangent to circle $S_{1}$, then it is als... | 2.30. Let's draw a line $l_{1}$ tangent to $S_{1}$ at point $A_{1}$. The line $K_{1} K_{2}$ is tangent to $S_{1}$ if and only if $\angle\left(K_{1} K_{2}, K_{1} A_{1}\right)=\angle\left(K_{1} A_{1}, l_{1}\right)$. It is also clear that $\angle\left(K_{1} A_{1}, l_{1}\right)=\angle\left(A_{1} B, l_{1}\right)=\angle\left... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,740 |
2.31. Isosceles trapezoids \(ABCD\) and \(A_1B_1C_1D_1\) are inscribed in a circle with their respective parallel sides. Prove that \(AC = A_1C_1\). | 2.31. Chords $A C$ and $A_{1} C_{1}$ subtend equal angles $A B C$ and $A_{1} B_{1} C_{1}$, therefore $A C=A_{1} C_{1}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,741 |
2.32. From a point $M$ moving along a circle, perpendiculars $M P$ and $M Q$ are dropped to the diameters $A B$ and $C D$. Prove that the length of the segment $P Q$ does not depend on the position of the point $M$. | 2.32. Let the center of the circle be denoted by $O$. Points $P$ and $Q$ lie on the circle with diameter $O M$, i.e., points $O, P, Q$ and $M$ lie on a circle of constant radius $R / 2$. In this case, either $\angle P O Q = \angle A O D$, or $\angle P O Q = \angle B O D = 180^{\circ} - \angle A O D$, i.e., the length o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,742 |
2.33. In triangle $ABC$, angle $B$ is equal to $60^{\circ}$, the bisectors $AD$ and $CE$ intersect at point $O$. Prove that $OD=OE$. | 2.33. Since $\angle A O C=90^{\circ}+\angle B / 2$ (see problem 5.3), then $\angle E B D+\angle E O D=90^{\circ}+$ $+3 \angle B / 2=180^{\circ}$, which means that quadrilateral $B E O D$ is cyclic. The chords $E O$ and $O D$ subtend equal angles $E B O$ and $O B D$, therefore $E O=O D$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,743 |
2.34. In triangle $ABC$, the angles at vertices $B$ and $C$ are $40^{\circ}$; $BD$ is the angle bisector of angle $B$. Prove that $BD + DA = BC$. | 2.34. Let's take a point $Q$ on the extension of segment $B D$ beyond point $D$ such that $\angle A C Q=40^{\circ}$. Let $P$ be the point of intersection of lines $A B$ and $Q C$. Then $\angle B P C=60^{\circ}$ and $D$ is the point of intersection of the angle bisectors of triangle $B C P$. According to problem $2.33, ... | BD+DA=BC | Geometry | proof | Yes | Yes | olympiads | false | 26,744 |
2.35. On the chord $A B$ of the circle $S$ with center $O$, a point $C$ is taken. The circumcircle of triangle $A O C$ intersects circle $S$ at point $D$. Prove that $B C = C D$. | 2.35. It is sufficient to check that the exterior angle $ACD$ of triangle $BCD$ is twice the angle at vertex $B$. It is clear that $\angle ACD = \angle AOD = 2 \angle ABD$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,745 |
2.36. Inside the square $A B C D$, a point $M$ is chosen such that $\angle M A C = \angle M C D = \alpha$. Find the measure of angle $A B M$. | 2.36. If point $M$ lies inside triangle $A B C$, then $\angle M A C < 45^{\circ} < \angle M C D$. It is also easy to verify that point $M$ cannot lie on the sides of triangles $A B C$ and $A C D$, so it must lie inside triangle $A C D$. In this case, $\angle A M C = 180^{\circ} - \angle M A C - (45^{\circ} - \angle M C... | 90-2\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,746 |
2.37. Vertices $A$ and $B$ of the equilateral triangle $ABC$ lie on the circle $S$, and vertex $C$ is inside this circle. Point $D$ lies on the circle $S$, and $BD = AB$. The line $CD$ intersects $S$ at point $E$. Prove that the length of the segment $EC$ is equal to the radius of the circle $S$. | 2.37. Let $O$ be the center of the circle $S$. Point $B$ is the center of the circumscribed circle of triangle $A C D$, so $\angle C D A = \angle A B C / 2 = 30^{\circ}$, and therefore, $\angle E O A = 2 \angle E D A = 60^{\circ}$, which means triangle $E O A$ is equilateral. Moreover, $\angle A E C = \angle A E D = \a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,747 |
2.39. From an arbitrary point $M$ on the leg $BC$ of a right triangle $ABC$, a perpendicular $MN$ is dropped to the hypotenuse $AB$. Prove that $\angle MAN = \angle MCN$.
保留源文本的换行和格式，直接输出翻译结果如下：
2.39. From an arbitrary point $M$ on the leg $BC$ of a right triangle $ABC$, a perpendicular $MN$ is dropped to the hypoten... | 2.39. Points $N$ and $C$ lie on a circle with diameter $A M$. Angles $M A N$ and $M C N$ subtend the same arc, so they are equal. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,749 |
2.40. The diagonals of trapezoid $ABCD$ with bases $AD$ and $BC$ intersect at point $O$; points $B'$ and $C'$ are symmetric to vertices $B$ and $C$ with respect to the bisector of angle $BOC$. Prove that $\angle C'AC = \angle B'DB$. | 2.40. With symmetry relative to the bisector of angle $B O C$, lines $A C$ and $D B$ transform into each other, so we need to prove that $\angle C^{\prime} A B^{\prime}=\angle B^{\prime} D C^{\prime}$. Since $B O=B^{\prime} O, C O=C^{\prime} O$ and $A O: D O=C O: B O$, then $A O \cdot B^{\prime} O=D O \cdot C^{\prime} ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,750 |
2.41. The extensions of sides $A B$ and $C D$ of the inscribed quadrilateral $A B C D$ intersect at point $P$, and the extensions of sides $B C$ and $A D$ intersect at point $Q$. Prove that the points of intersection of the angle bisectors of $\angle A Q B$ and $\angle B P C$ with the sides of the quadrilateral are the... | 2.41. Let's denote the points of intersection and angles as shown in Fig. 2.1. It is sufficient to check that \( x = 90^{\circ} \). The angles of quadrilateral \( B M R N \) are \( 180^{\circ} - \varphi, \alpha + \varphi, \beta + \varphi \) and \( x \), so the equality \( x = 90^{\circ} \) is equivalent to the equality... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,751 |
2.42*. The inscribed circle touches the sides $AB$ and $AC$ of triangle $ABC$ at points $M$ and $N$. Let $P$ be the point of intersection of line $MN$ and the bisector of angle $B$ (or its extension). Prove that:
a) $\angle B P C=90^{\circ}$;
b) $S_{A B P}: S_{A B C}=1: 2$. | 2.42. a) It is sufficient to prove that if $P_{1}$ is a point on the bisector of angle $B$ (or its extension), from which the segment $B C$ is seen at an angle of $90^{\circ}$, then $P_{1}$ lies on the line $M N$. Points $P_{1}$ and $N$ lie on the circle with diameter $C O$, where $O$ is the intersection of the bisecto... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,752 |
2.43*. Inside the quadrilateral $A B C D$, a point $M$ is taken such that $A B M D$ is a parallelogram. Prove that if $\angle C B M = \angle C D M$, then $\angle A C D = \angle B C M$. | 2.43. Let's take point $N$ such that $B N \| M C$ and $N C \| B M$. Then $N A \| C D, \angle N C B = \angle C B M = \angle C D M = \angle N A B$, i.e., points $A, B, N$ and $C$ lie on the same circle. Therefore, $\angle A C D = \angle N A C = \angle N B C = \angle B C M$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,753 |
2.44. The lines $A P, B P$ and $C P$ intersect the circumcircle of triangle $A B C$ at points $A_{1}, B_{1}$ and $C_{1}$. Points $A_{2}, B_{2}$ and $C_{2}$ are taken on the lines $B C, C A$ and $A B$ such that $\angle\left(P A_{2}, B C\right)=\angle\left(P B_{2}, C A\right)=\angle\left(P C_{2}, A B\right)$. Prove that ... | 2.44. Points $A_{2}, B_{2}, C$ and $P$ lie on the same circle, so $\angle\left(A_{2} B_{2}, B_{2} P\right)=\angle\left(A_{2} C, C P\right)=\angle(B C, C P)$. Similarly, $\angle\left(B_{2} P, B_{2} C_{2}\right)=$ $=\angle(A P, B P)$. Therefore, $\angle\left(A_{2} B_{2}, B_{2} C_{2}\right)=\angle(B C, C P)+\angle(A P, A ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,754 |
2.45*. A rectangle $A B C D$ is circumscribed around an equilateral triangle $A P Q$, with points $P$ and $Q$ lying on sides $B C$ and $C D$ respectively; $P^{\prime}$ and $Q^{\prime}$ are the midpoints of sides $A P$ and $A Q$. Prove that triangles $B Q^{\prime} C$ and $C P^{\prime} D$ are equilateral.
2.46* ${ }^{*}... | 2.45. Points $Q^{\prime}$ and $C$ lie on the circle with diameter $P Q$, so $\angle Q^{\prime} C Q=$ $=\angle Q^{\prime} P Q=30^{\circ}$. Therefore, $\angle B C Q^{\prime}=60^{\circ}$. Similarly, $\angle C B Q^{\prime}=60^{\circ}$, which means triangle $B Q^{\prime} C$ is equilateral. Similarly, triangle $C P^{\prime} ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,755 |
2.48*. Triangles $A B C$ and $A_{1} B_{1} C_{1}$ have corresponding parallel sides, and sides $A B$ and $A_{1} B_{1}$ lie on the same line. Prove that the line connecting the points of intersection of the circumcircles of triangles $A_{1} B C$ and $A B_{1} C$ contains the point $C_{1}$. | 2.48. Let $D$ be the second intersection point of the circumcircles of triangles $A_{1} B C$ and $A B_{1} C$. Then $\angle(A C, C D)=\angle\left(A B_{1}, B_{1} D\right)$ and $\angle(D C, C B)=$ $=\angle\left(D A_{1}, A_{1} B\right)$. Therefore, $\angle\left(A_{1} C_{1}, C_{1} B_{1}\right)=\angle(A C, C B)=\angle(A C, C... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,756 |
2.49*. In triangle $A B C$, the altitudes $A A_{1}, B B_{1}$, and $C C_{1}$ are drawn. The line $K L$ is parallel to $C C_{1}$, with points $K$ and $L$ lying on the lines $B C$ and $B_{1} C_{1}$, respectively. Prove that the center of the circumcircle of triangle $A_{1} K L$ lies on the line $A C$. | 2.49. Let point $M$ be symmetric to point $A_{1}$ with respect to the line $A C$. According to problem 1.58, point $M$ lies on the line $B_{1} C_{1}$. Therefore, $\angle\left(L M, M A_{1}\right)=$ $=\angle\left(C_{1} B_{1}, B_{1} A\right)=\angle\left(C_{1} C, C B\right)=\angle\left(L K, K A_{1}\right)$, i.e., point $M$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,757 |
2.50*. Through the point $O$ of intersection of the angle bisectors of triangle $ABC$, a line $MN$ is drawn perpendicular to $CO$, with $M$ and $N$ lying on sides $AC$ and $BC$ respectively. The lines $AO$ and $BO$ intersect the circumcircle of triangle $ABC$ at points $A'$ and $B'$. Prove that the intersection point o... | 2.50. Let $P Q$ be a diameter perpendicular to $A B$, with $Q$ and $C$ lying on the same side of $A B$; $L$ be the intersection point of the line $Q O$ with the circumscribed circle; $M^{\prime}$ and $N^{\prime}$ be the intersection points of the lines $L B^{\prime}$ and $L A^{\prime}$ with the sides $A C$ and $B C$. I... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,758 |
2.51. Points $A, B, C$, and $D$ are taken on a circle. Lines $A B$ and $C D$ intersect at point $M$. Prove that $A C \cdot A D / A M = B C \cdot B D / B M$. | 2.51. Since $\triangle A D M \sim \triangle C B M$ and $\triangle A C M \sim \triangle D B M$, then $A D: C B=$ $=D M: B M$ and $A C: D B=A M: D M$. It remains to multiply these equalities. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,759 |
2.52. On a circle, points $A, B$, and $C$ are given, with point $B$ being farther from the line $l$, tangent to the circle at point $A$, than $C$. The line $A C$ intersects the line through point $B$ parallel to $l$ at point $D$. Prove that $A B^{2}=A C \cdot A D$. | 2.52. Let $D_{1}$ be the point of intersection of the line $B D$ with the circle, different from point $B$. Then $\smile A B=\smile A D_{1}$, so $\angle A C B=\angle A D_{1} B=\angle A B D_{1}$. Triangles $A C B$ and $A B D$ have a common angle $A$ and, in addition, $\angle A C B=\angle A B D$, so $\triangle A C B \sim... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,760 |
2.53. The line $l$ is tangent to the circle with diameter $A B$ at point $C; M$ and $N$ are the projections of points $A$ and $B$ onto the line $l, D$ is the projection of point $C$ onto $A B$. Prove that $C D^{2}=A M \cdot B N$. | 2.53. Let $O$ be the center of the circle. Since $\angle M A C = \angle A C O = \angle C A O$, then $\triangle A M C = \triangle A D C$. Similarly, $\triangle C D B = \triangle C N B$. Since $\triangle A C D \sim \triangle C D B$, then $C D^{2} = A D \cdot D B = A M \cdot N B$. | CD^{2}=AM\cdotBN | Geometry | proof | Yes | Yes | olympiads | false | 26,761 |
2.54. In triangle $ABC$, the altitude $AH$ is drawn, and perpendiculars $BB_1$ and $CC_1$ are dropped from vertices $B$ and $C$ to a line passing through point $A$. Prove that $\triangle ABC \sim \triangle HB_1C_1$. | 2.54. Points $B_{1}$ and $H$ lie on the circle with diameter $A B$, therefore $\angle(A B, B C)=\angle(A B, B H)=\angle\left(A B_{1}, B_{1} H\right)=\angle\left(B_{1} C_{1}, B_{1} H\right)$. Similarly, we have $\angle(A C, B C)=\angle\left(B_{1} C_{1}, C_{1} H\right)$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,762 |
2.55. On the arc $BC$ of the circumcircle of an equilateral triangle $ABC$, a point $P$ is taken arbitrarily. The segments $AP$ and $BC$ intersect at point $Q$. Prove that $1 / PQ = 1 / PB + 1 / PC$. | 2.55. On the extension of segment $B P$ beyond point $P$, take point $D$ such that $P D = C P$. Then triangle $C D P$ is equilateral and $C D \| Q P$. Therefore, $B P: P Q = B D: D C = (B P + C P): C P$, i.e., $\frac{1}{P Q} = \frac{1}{C P} + \frac{1}{B P}$. | \frac{1}{PQ}=\frac{1}{CP}+\frac{1}{BP} | Geometry | proof | Yes | Yes | olympiads | false | 26,763 |
2.56. On the sides $B C$ and $C D$ of the square $A B C D$, points $E$ and $F$ are taken such that $\angle E A F=45^{\circ}$. Segments $A E$ and $A F$ intersect the diagonal $B D$ at points $P$ and $Q$. Prove that $S_{A E F} / S_{A P Q}=2$. | 2.56. The segment $Q E$ is seen from points $A$ and $B$ at an angle of $45^{\circ}$, so the quadrilateral $A B E Q$ is cyclic. Since $\angle A B E=90^{\circ}$, then $\angle A Q E=90^{\circ}$. Therefore, triangle $A Q E$ is a right isosceles triangle and $A E / A Q=\sqrt{2}$. Similarly, $A F / A P=\sqrt{2}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,764 |
2.57. A line passing through vertex $C$ of isosceles triangle $ABC$ intersects the base $AB$ at point $M$ and the circumscribed circle at point $N$. Prove that $CM \cdot CN = AC^2$ and $CM / CN = AM \cdot BM / (AN \cdot BN)$. | 2.57. Since $\angle A N C = \angle A B C = \angle C A B$, then $\triangle C A M \sim \triangle C N A$, which means $C A : C M = C N : C A$, i.e., $C M \cdot C N = A C^{2}$, and $A M : N A = C M : C A$. Similarly, $B M : N B = C M : C B$. Therefore, $A M \cdot B M / (A N \cdot B N) = C M^{2} / C A^{2} = C M^{2} / (C M \... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,765 |
2.58. Given a parallelogram $A B C D$ with an acute angle at vertex $A$. On the rays $A B$ and $C B$, points $H$ and $K$ are marked respectively such that $C H=B C$ and $A K=A B$. Prove that:
a) $D H=D K$;
b) $\triangle D K H \sim \triangle A B K$. | 2.58. Since $A K=A B=C D, A D=B C=C H$ and $\angle K A D=\angle D C H$, then $\triangle A D K=\triangle C H D$ and $D K=D H$. We will show that points $A, K, H, C$ and $D$ lie on the same circle. Describe a circle around triangle $A D C$. Draw a chord $C K_{1}$ parallel to $A D$ and a chord $A H_{1}$ parallel to $D C$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,766 |
2.59. a) The sides of an angle with vertex $C$ touch a circle at points $A$ and $B$. From point $P$, lying on the circle, perpendiculars $P A_{1}, P B_{1}$, and $P C_{1}$ are dropped to the lines $B C, C A$, and $A B$. Prove that $P C_{1}^{2} = P A_{1} \cdot P B_{1}$ and $P A_{1}: P B_{1} = P A^{2}: P B^{2}$.
b) From ... | 2.59. a) $\angle P B A_{1}=\angle P A C_{1}$ and $\angle P B C_{1}=\angle P A B_{1}$, so right triangles $P B A_{1}$ and $P A C_{1}$, $P A B_{1}$ and $P B C_{1}$ are similar, i.e., $P A_{1}: P B = P C_{1}: P A, P B_{1}: P A = P C_{1}: P B$. Multiplying these equalities, we get $P A_{1} \cdot P B_{1} = P C_{1}^{2}$, and... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,767 |
2.60. Pentagon $A B C D E$ is inscribed in a circle. The distances from point $E$ to the lines $A B, B C$ and $C D$ are $a, b$ and $c$ respectively. Find the distance from point $E$ to the line $A D$. | 2.60. Let $K, L, M$ and $N$ be the feet of the perpendiculars dropped from point $E$ to the lines $A B, B C, C D$ and $D A$. Points $K$ and $N$ lie on the circle with diameter $A E$, so $\angle(E K, K N)=\angle(E A, A N)$. Similarly, $\angle(E L, L M)=$ $=\angle(E C, C M)=\angle(E A, A N)$, hence $\angle(E K, K N)=\ang... | \frac{ac}{b} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,768 |
2.61. In triangle $A B C$, altitudes $A A_{1}, B B_{1}$, and $C C_{1}$ are drawn; $B_{2}$ and $C_{2}$ are the midpoints of altitudes $B B_{1}$ and $C C_{1}$. Prove that $\triangle A_{1} B_{2} C_{2} \sim \triangle A B C$. | 2.61. Let $H$ be the orthocenter, and $M$ be the midpoint of side $B C$. Points $A_{1}, B_{2}$, and $C_{2}$ lie on the circle with diameter $M H$, so $\angle\left(B_{2} A_{1}, A_{1} C_{2}\right)=$ $=\angle\left(B_{2} M, M C_{2}\right)=\angle(A C, A B)$. Moreover, $\angle\left(A_{1} B_{2}, B_{2} C_{2}\right)=\angle\left... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,769 |
2.62. On the altitudes of triangle $ABC$, points $A_{1}, B_{1}$, and $C_{1}$ are taken, dividing them in the ratio 2:1, counting from the vertex. Prove that $\triangle A_{1} B_{1} C_{1} \sim \triangle A B C$. | 2.62. Let $M$ be the point of intersection of the medians, $H$ be the point of intersection of the altitudes of triangle $ABC$. Points $A_{1}, B_{1}$, and $C_{1}$ are the projections of point $M$ onto the altitudes, so they lie on a circle with diameter $MH$. Therefore, $\angle\left(A_{1} B_{1}, B_{1} C_{1}\right)=$ $=... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,770 |
2.63*. Circle $S_{1}$ with diameter $A B$ intersects circle $S_{2}$ with center $A$ at points $C$ and $D$. A line through point $B$ intersects $S_{2}$ at point $M$, which lies inside $S_{1}$, and $S_{1}$ at point $N$. Prove that $M N^{2}=C N \cdot N D$. | 2.63. Let lines $B M$ and $D N$ intersect $S_{2}$ at points $L$ and $C_{1}$ respectively. We will prove that lines $D C_{1}$ and $C N$ are symmetric with respect to line $A N$. Since $B N \perp N A$, it is sufficient to check that $\angle C N B = \angle B N D$. But arcs $C B$ and $B D$ are equal. Arcs $C_{1} M$ and $C ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,771 |
2.64*. Through the midpoint $C$ of an arbitrary chord $A B$ of a circle, two chords $K L$ and $M N$ are drawn (points $K$ and $M$ lie on the same side of $A B$). Segments $K N$ and $M L$ intersect $A B$ at points $Q$ and $P$. Prove that $P C = Q C$. | 2.64. Drop perpendiculars $Q K_{1}$ and $Q N_{1}$ from point $Q$ to $K L$ and $N M$, and perpendiculars $P M_{1}$ and $P L_{1}$ from point $P$ to $N M$ and $K L$. Clearly, $\frac{Q C}{P C}=\frac{Q K_{1}}{P L_{1}}=$ $=\frac{Q N_{1}}{P M_{1}}$, i.e., $\frac{Q C^{2}}{P C^{2}}=\frac{Q K_{1} \cdot Q N_{1}}{P L_{1} \cdot P M... | QC=PC | Geometry | proof | Yes | Yes | olympiads | false | 26,772 |
2.66. In triangle $A B C$, sides $A C$ and $B C$ are not equal. Prove that the bisector of angle $C$ bisects the angle between the median and the altitude drawn from this vertex if and only if $\angle C=90^{\circ}$. | 2.66. Let $O$ be the center of the circumscribed circle of a triangle, $M$ the midpoint of side $AB$, $H$ the foot of the altitude $CH$, and $D$ the midpoint of the arc defined by points $A$ and $B$ that does not contain point $C$. Since $OD \parallel CH$, we have $\angle DCH = \angle MDC$. The bisector divides the ang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,774 |
2.68. Prove that in any triangle $ABC$ the bisector $AE$ lies between the median $AM$ and the altitude $AH$. | 2.68. Let $D$ be the point where the line $A E$ intersects the circumcircle. The point $D$ is the midpoint of the arc $B C$. Therefore, $M D \| A H$, and points $A$ and $D$ lie on opposite sides of the line $M H$. Consequently, the point $E$ lies on the segment $M H$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,776 |
2.69. Given a triangle $A B C$. On its side $A B$, a point $P$ is chosen and lines $P M$ and $P N$ are drawn parallel to $A C$ and $B C$ respectively (points $M$ and $N$ lie on sides $B C$ and $A C$); $Q$ is the point of intersection of the circumcircles of triangles $A P N$ and $B P M$. Prove that all lines $P Q$ pass... | 2.69. It is clear that $\angle(A Q, Q P)=\angle(A N, N P)=\angle(P M, M B)=\angle(Q P, Q B)$. Therefore, point $Q$ lies on the circle from which segment $A B$ is seen at an angle of $2 \angle(A C, C B)$, and line $Q P$ bisects the arc $A B$ of this circle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,777 |
2.70. The continuation of the bisector $A D$ of an acute-angled triangle $A B C$ intersects the circumscribed circle at point $E$. Perpendiculars from point $D$ to the sides $A B$ and $A C$ are dropped to points $D P$ and $D Q$. Prove that $S_{A B C} = S_{A P E Q}$.
## §8. Inscribed Quadrilateral with Perpendicular Di... | 2.70. Points $P$ and $Q$ lie on the circle with diameter $A D$; this circle intersects side $B C$ at point $F$ ( $F$ does not coincide with $D$ if $A B \neq A C$ ). It is clear that $\angle(F C, C E)=\angle(B A, A E)=\angle(D A, A Q)=\angle(D F, F Q)$, i.e., $E C \| F Q$. Similarly, $B E \| F P$. To complete the proof,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,778 |
2.74. Perpendiculars are dropped from vertices $A$ and $B$ to $CD$, intersecting lines $BD$ and $AC$ at points $K$ and $L$ respectively. Prove that $AKLB$ is a rhombus. | 2.74. The acute angles $B L P$ and $B D C$ have perpendicular sides respectively, so they are equal. Therefore, $\angle B L P = \angle B D C = \angle B A P$. Moreover, $A K \| B L$ and $A L \perp B K$. Therefore, $A K L B$ is a rhombus. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,782 |
2.75. Prove that the area of quadrilateral $A B C D$ is equal to $(A B \cdot C D + B C \cdot A D) / 2$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 2.75. Let's take a point $D^{\prime}$ on the circumscribed circle such that $D D^{\prime} \| A C$. Since $D D^{\prime} \perp B D$, then $B D^{\prime}$ is a diameter, which means $\angle D^{\prime} A B=\angle D^{\prime} C B=90^{\circ}$. Therefore, $S_{A B C D}=S_{A B C D^{\prime}}=\left(A D^{\prime} \cdot A B+B C \cdot ... | Number Theory | proof | Yes | Yes | olympiads | false | 26,783 | |
2.79. a) Tangents are drawn through the vertices $A, B, C$ and $D$ of a cyclic quadrilateral. Prove that the quadrilateral formed by these tangents is cyclic.
b) The quadrilateral $K L M N$ is both cyclic and tangential; $A$ and $B$ are the points of tangency of the inscribed circle with the sides $K L$ and $L M$. Pro... | 2.79. a) It should be noted that since points $A, B, C$ and $D$ divide the circle into arcs less than $180^{\circ}$, the constructed quadrilateral contains this circle. The angle $\varphi$ between the tangents drawn through points $A$ and $B$ is $180^{\circ}-\angle A O B$, and the angle $\psi$ between the tangents draw... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,787 |
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