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131. If the circles inscribed in the triangles into which a quadrilateral is divided by one of its diagonals touch each other, then the circles inscribed in the triangles into which the quadrilateral is divided by the other diagonal also touch each other. | 131. Establish that if the specified circles touch, then a circle can be inscribed in the quadrilateral, and then prove the converse by contradiction. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,707 |
132. Through three pairs of vertices of a triangle, three circles are drawn such that the arcs lying outside the triangle accommodate angles whose sum is $180^{\circ}$. Prove that these circles pass through one point. | 132. Determine that the intersection point of two circles lies on a third circle. Use the properties of angles in cyclic quadrilaterals. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,708 |
135. In circle $O$, triangle $A B C$ is inscribed, and is circumscribed about circle $O^{\prime}$. Prove that if point $M$ is the midpoint of arc $A B$ not containing point $C$, then $M A = M B = M O^{\prime}$. | 135. Establish the equality of angles $M B O^{\prime}$ and $M O^{\prime} B$. Use the properties of inscribed angles and the property of an exterior angle of a triangle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,711 |
136. A circle is circumscribed around triangle $A B C$. Prove that the point $D$, diametrically opposite to point $C$, is symmetric to the orthocenter of the triangle with respect to the midpoint of side $A B$. | 136. Note that the distance from the orthocenter $H$ of a triangle to its vertex $C$ is equal to twice the distance from the center of the circumscribed circle to the corresponding side. Connect the center of the circumscribed circle with the midpoint of segment $\mathrm{CH}$. Use the theorem about the midline of a tri... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,712 |
138. Prove that the circles passing through each two vertices of a triangle and its orthocenter are equal to the circle circumscribed about the triangle. | 138. Establish that each of these circles is symmetric to the circumscribed circle with respect to the corresponding side of the triangle. Use the result of problem № 137. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,714 |
140. On the side $AB$ of triangle $ABC$, a semicircle is constructed with $AB$ as its diameter, intersecting sides $AC$ and $BC$ or their extensions at points $M$ and $N$ respectively. Prove that the tangents drawn to the semicircle at points $M$ and $N$ intersect on the altitude $\mathrm{CH}_{3}$ of the triangle or on... | 140. Prove that $\angle M K N=2 \angle C$ (or $\angle M K N+2 \angle C=360^{\circ}$). Given that $M K=N K$, establish that $K$ is the center of the circumcircle of triangle $M N C$. On this circle, point $H$ also lies. Since $H C$ is the diameter of the circle, prove that point $K$ lies on $C H$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,716 |
141. A line passing through the center of the circumscribed circle and the orthocenter of triangle $ABC$ cuts off equal segments $CA_1, CB_1$ from its sides $CA$ and $CB$. Prove that angle $C$ is $60^{\circ}$.
## § 9. Geometric Loci
| 141. Drop a perpendicular $O M_{2}$ from the center $O$ of the circumscribed circle onto $A C$. Triangles $C M_{2} O$ and $C H H_{1}$ are equal, and $C H_{1}=\frac{1}{2} A C$, which means that $\angle C=60^{\circ}$. | 60 | Geometry | proof | Yes | Yes | olympiads | false | 43,717 |
142. Through the ends of segments $A B$ and $B C$, lying on the same line, equal circles are drawn, intersecting again at point $M$. Prove that the geometric locus of points $M$ is a perpendicular (with an excluded point) to the segment $A C$, passing through its midpoint. | 142. Establish that triangle $A M C (A M B)$ is isosceles and point $M$ projects to a constant point on line $A C$. Consider two cases: 1) point $B$ lies between $A$ and $C$, 2) point $C$ lies between $A$ and $B$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,718 |
143. The vertices of the acute angles of a right triangle slide along the sides of the right angle. Prove that the geometric locus of the vertex of the right angle of the triangle is a pair of segments corresponding to the two possible positions of the triangle relative to the sides of the angle. | 143. If the vertex of the right angle of the sliding triangle and the vertex of the given right angle lie on opposite sides of the hypotenuse, then the sought geometric locus is a segment that forms angles with the sides of the angle equal to the angles of the right triangle; the ends of the segment are at distances fr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,719 |
144. Perpendiculars are erected at some points $A$ and $B$ on the sides of angle $C$, intersecting at point $M$. Prove that the geometric locus of points $M$ under the condition that the perpendiculars to the sides of the angle are erected at points which are the endpoints of segments parallel to $A B$, is a ray. | 144. Describe a circle around the quadrilateral $A M B C$. Establish that the segment $C M_{l}$ forms constant angles with the sides of the given angle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,720 |
145. Given an equilateral triangle $A B C$. Prove that the geometric locus of points $M$ for which the segments cut off on the lines $A M$ and $B M$ by the sides of the triangle or their extensions are equal, is a straight line and a circle. | 145. Notice that the lines $M A$ and $M B$ either intersect on the bisector of angle $C$, or $\angle A M B=120^{\circ}$, or $\angle A M B=60^{\circ}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,721 |
146. Two circles touching a given line at given points touch each other. Prove that the geometric locus of the points of contact of the circles is a circle with two excluded points. | 146. Prove that from the point of tangency of circles, the segment of their common tangent is seen at a right angle. The sought geometric locus of points is a circle having as its diameter the segment with endpoints at the given two points, which do not belong to the geometric locus. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,722 |
147. On each of two intersecting circles $O_{1}$ and $O_{2}$ at points $A$ and $B$, points $S_{1}$ and $S_{2}$ are given, respectively. A variable line $g_{1}$ passing through point $S_{1}$ intersects circle $O_{1}$ at point $P_{1}$, and line $A P$ intersects circle $O_{2}$ at point $P_{2}$. Prove that the geometric lo... | 147. Construct two points of the geometric locus and consider two pairs of equal inscribed angles. Use the criterion for the concyclicity of four points. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,723 |
148. A segment of constant length $a$ slides with its ends along the sides of an angle. Prove that the geometric locus of the points of intersection of the perpendiculars erected to the sides of the angle at the ends of the segment is an arc of a circle. | 148. To establish that the intersection point of the perpendiculars at the ends of a segment is located at a constant distance from the vertex of the angle, equal to the diameter of a variable circle of constant radius determined by the given angle and the given segment. The geometric locus of points is an arc of a cir... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,724 |
150. Through the ends of the diameter $A B$ and the chord $C D$ of the circle $O$, lines $A C$ and $B D$ are drawn, intersecting at point $M$. Prove that: 1) the geometric locus of points $M$ when the diameter $A B$ makes a full revolution is a circle; 2) the geometric locus of points $M$ when the ends of the chord $C ... | 150. 1) Establish that the angle $C M D$ takes two values, the sum of which is $180^{\circ}$. 2) A circle passes through the ends of the diameter $A B$. If we swap the labels of the ends of the chord, we obtain a second circle, symmetric to the first relative to the diameter $A B$. Both these circles are tangent to the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,726 |
151. On the side $AB$ (or its extension) of triangle $ABC$, a point $K$ is given, through which an arbitrary secant is drawn, intersecting sides $AC$ and $BC$ at points $M$ and $N$, respectively.
The circumcircles of triangles $AKM$ and $BKN$ intersect at point $P$. Prove that the geometric locus of points $P$ is the ... | 151. Prove that $\angle A P B + \angle C = 180^{\circ}$, or $\angle A P B = \angle C$. Consider the quadrilaterals $A M P K$ and $B N K P$, inscribed in the given circles. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,727 |
152. On the chord $AB$ of the circle $O$, as on a diameter, a second circle $O_{1}$ is described. Through the point $A$, a secant is drawn, intersecting the circles $O_{1}$ and $O$ at points $M_{i}$ and $N_{i}$, respectively. Prove that the geometric locus of points symmetric to $N_{i}$ with respect to points $M_{i}$, ... | 152. The reflected point $P$ has the property that $\angle A P B = \angle A N B$, or $\angle A P B + \angle A N B = 180^{\circ}$. The desired geometric locus of points is a circle symmetric to the given one with respect to the line $A B$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,728 |
153. On the sides of angle $C$, two points $A$ and $B$ are given such that $A C = B C$. Prove that the geometric locus of points $M$, located inside the angle and for which the ray $M C$ is the bisector of angle $A M B$, is an open arc of a circle and the bisector of angle $C$ (excluding the vertex). | 153. The geometric locus of points $M$ is the bisector of angle $C$ and the arc of the circle subtended by the chord $A B$ and containing the angle $\left(180^{\circ}-\angle C\right)$. To establish this, consider the case,
.
## § 10. Mixed Problems | 155. Let two orthogonal circles $O_{1}$ and $O_{2}$ intersect at point $P$ (Fig. 12). Suppose $\angle O_{1} C A = \angle O_{1} A C = \varphi, \angle O_{2} C B = \angle O_{2} B C = \psi, \angle C O_{1} P = \alpha$, and $\angle O_{1} C_{2} = \angle O_{1} P O_{2} = 90^{\circ}$. Therefore, $\angle A O_{1} P = 180^{\circ} -... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,731 |
156. From a point \( P \) located inside an equilateral triangle \( ABC \), the side \( AB \) is seen at an angle of \( 150^\circ \). Prove that a right triangle can be constructed from the segments \( PA, PB, \) and \( PC \). | 156. Rotate the segment $C P$ (Fig. 13) around point $C$ by $60^{\circ}$ to the position $C P_{1}$. Let $\angle P P_{1} B=\alpha, \angle P_{1} P B=\beta$. Then $\angle A P C=$ $=\angle C P_{1} B=60^{\circ}+\alpha, \angle B P C=60^{\circ}+\beta$. Therefore, $150^{\circ}+60^{\circ}+$ $+\alpha+60^{\circ}+\beta=360^{\circ}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,732 |
158. The height and median of a triangle, drawn from the same vertex, divide the angle at this vertex into three equal parts. Prove that the given triangle is a right triangle. | 158. Let $\angle A C H_{3}=\angle H_{3} C M_{3}=\angle M_{3} C B$ (Fig. 14). Notice that $B M_{3}=2 M_{3} H_{3}$. Reflect vertex $C$ over $A B$ to point $C_{1}$ and prove that in triangle $B C C_{1}$, point $M_{3}$ is both the centroid and the incenter. Therefore, triangle $B C C_{1}$ is equilateral and $\angle A C B=9... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,734 |
159. On the side of a triangle or its extension, a point is taken from which perpendiculars are dropped to the other two sides. Prove that the distance between the bases of these perpendiculars will be the smallest if the taken point is the base of the triangle's altitude. | 159. First solution. From point $M$ of side $A B$, perpendiculars $M P$ and $M Q$ are dropped to the other two sides, which are doubled (Fig. 15). It is required to establish that $C P_{1}=C Q_{1}=M C, P_{1} Q_{1}=2 P Q$ and $\angle P_{1} C Q_{1}=2 \angle A C B$. Consequently, the base $P_{1} Q_{1}$ of the isosceles tr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,735 |
161. The difference between angles $B$ and $A$ of triangle $A E C$ is $90^{\circ}$. From the base $H_{3}$ of the altitude $C H_{3}$, perpendiculars $H_{3} M$ and $H_{3} N$ are dropped to sides $A C$ and $E C$. Prove that line $M N$ is perpendicular to line $A B$. | 161. First method. Prove that $\angle B H_{3} N=\angle A$ and therefore the line $H_{3} N$ intersects the side $A C$ at its midpoint $S$. Establish that triangle $\mathrm{CH}_{3} S$ is isosceles, and quadrilateral $\mathrm{MNH}_{3} \mathrm{C}$ is cyclic.
Second method. At vertex $B$ of the obtuse angle of triangle $A ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,737 |
162. Through the vertices $A$ and $B$ of the base $AB$ of an isosceles triangle $ABC$ with an angle of $20^{\circ}$ at vertex $C$, two secants are drawn at angles of $50^{\circ}$ and $60^{\circ}$ to the base, respectively, intersecting sides $BC$ and $AC$ at points $A_{1}$ and $B_{1}$. Prove that the angle $A_{1} B_{1}... | 162. Draw a secant through vertex $A$ at an angle of $60^{\circ}$ to the base $A B$ (Fig. 16), which intersects segment $B B_{1}$ and side $B C$ at points $M$ and $N$, respectively. Establish that triangles $1 M B$ and $M N B_{1}$ are equilateral, and triangle $A_{1} M B$ is isosceles $\left(M B=A B=A_{1} B\right)$. Fr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,738 |
164. Tangents are drawn to circle $O$ at points $A$ and $B$. Through an arbitrary point $M$ on chord $AB$, a secant is drawn perpendicular to line $OM$. Prove that the segments of the secant, enclosed between the circle and the tangents, are equal to each other. | 164. Let a secant meet $B C$ at point $P$, and the extension of $A C$ at point $Q$. Consider two cyclic quadrilaterals $O M A Q$ and $O M P B$. Prove the equality of angles $Q O M$ and $M O P$, using the equality of angles $C A B$ and $C B A$. From this, $Q M = M P$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,740 |
165. From the vertex $C$ of triangle $A B C$, the altitude $C H_{3}$, the bisector $C L_{3}$, and the median $C M_{3}$ are drawn. Prove that if the angles $A C H_{2}$, $H_{3} C L_{3}$, $L_{3} C M_{3}$, $M_{3} C B$ are equal, then triangle $A B C$ is a right triangle with an acute angle of $22.5^{\circ}$. | 165. Establish that the equality of angles $A C H_{3}$ and $M_{3} C B$ implies the perpendicularity of sides $A C$ and $B C$. Consequently, $\angle B=22.5^{\circ}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,741 |
167. At the end of the chord $AB$ of circle $O$, a tangent is drawn, and from point $B$, a perpendicular $BM$ is dropped to this tangent, intersecting the circle again at point $C$. Prove that the center $O$, the point $N$ dividing the chord $AB$ in the ratio $1:2$, and the point $C'$, symmetric to point $C$ with respe... | 167. Prove that triangle $B A C^{\prime}$ is a right triangle. Drop a perpendicular from the center $O$ to the chord $A B$, meeting $B C^{\prime}$ at $S$, and prove that point $S$ is the midpoint of $B C^{\prime}$. Draw the midline in triangle $A C^{\prime} N$ parallel to the leg $A C^{\prime}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,743 |
168. Circle $O$ touches line $m$, to which a perpendicular is erected at its point $M$, intersecting the circle at points $A$ and $B$. Prove that if point $A^{\prime}$ is symmetric to point $A$ with respect to point $M$, then segment $A^{\prime} B$ is equal to the diameter of the circle and is seen from the point of ta... | 168. Establish that $\angle A^{\prime} T M=\angle T B A^{\prime}$, from which it follows that $T A^{\prime} \perp T B$ ( $T$ - the point of tangency). Draw the diameter $T T_{1}$ through point $T$ and prove the parallelism of $B T_{1}$ and $A^{\prime} T$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,744 |
169. Prove that if a line passes through the vertex of a triangle perpendicular to the median drawn from the same vertex, then the sum of the distances from the other two vertices of the triangle to this line is the greatest. | 169. Reflect the triangle and the secant line relative to the midpoint of the side and consider the resulting rectangle. On two sides of the parallelogram, as diameters, describe circles and establish that the segment of the secant, passing through the vertex of the triangle and enclosed within the circles, will be max... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,745 |
170. In the plane of triangle $A B C$, a point $M$ is given, through which lines perpendicular to $M A, M B$, and $M C$ are drawn. Prove that the points $A_{1}, B_{1}, C_{1}$ - the points of intersection of these lines with the corresponding sides of the triangle, either all lie on the extensions of the sides, or only ... | 170. Consider three cases corresponding to different possible positions of point $M$ relative to triangle $A B C$. For example, if point $M$ lies inside the triangle, then two angles at this point are necessarily obtuse, while the third can be acute, right, or obtuse. Let's determine the validity of the theorem for eac... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,746 |
171. In the plane of an equilateral triangle $A B C$, a point $M$ is given. Prove that a triangle can be constructed from the segments $M A, M B, M C$, which degenerates only for points located on the circumcircle of the triangle. | 171. Rotate the segment $M B$ around point $M$ by $60^{\circ}$, forming an equilateral triangle $M B B_{1}$ and two equal triangles $A B M$ and $B B_{1} C$. Establish that if point $M$ lies on the circumcircle of the given triangle, then triangle $M B_{1} C$ degenerates. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,747 |
172. Triangle $A B C$ is inscribed in a circle. Through an arbitrary point $P$ of the circle, lines parallel to the sides of the triangle are drawn and intersect the circle again at points $C_{1}, A_{1}, B_{1}$. Prove that: a) triangles $A B C$ and $A_{1} B_{1} C_{1}$ are equal, b) the lines $A A_{1}, B B_{1}, C C_{1}$... | 172. Angles $A_{1} P C_{1}$ and $A B C$ are equal (or their sum is $180^{\circ}$). Therefore, $A C=A_{1} C_{1}$. From the equalities $A_{1} C=B P$ and $B P=$ $=A C_{1}$, it follows that $A_{1} C=A C_{1}$. Thus, $A A_{1} \| C C_{1}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,748 |
174. On a circle, three points $A, B, C$ are given. The last point is reflected with respect to the midpoint of segment $A B$. The resulting point $C_{1}$ is connected to point $D$, which is diametrically opposite to point $C$. Prove the perpendicularity of lines $A B$ and $C_{1} D$. | 174. Consider triangle $C C_{1} D$ and establish that the midline $O M_{3}$ of this triangle is perpendicular to the chord $A B$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,750 |
175. Given a circle and points $A$ and $B$, from which tangents are drawn to the circle. Prove that if the line passing through the points of tangency of two tangents, not originating from the same point, bisects the segment $A B$ or is parallel to it, then the segments of the tangents, bounded by the given points and ... | 175. First method. Drop a perpendicular from the center to the chord connecting the points of tangency, and consider it as the axis of symmetry for the tangents. Reflect point $A$ with respect to this axis and use the theorem about the midline of a triangle (Fig. 17).
Second method. Through one of the given points, dr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,751 |
176. Two circles $O$ and $O_{1}$ of equal radii intersect at points $A$ and $B$. Through point $A$, their common secant $C A D$ is drawn, and from point $B$, a perpendicular is dropped to this secant, intersecting the circles at points $E$ and $F$. Prove that: a) points $C, E, D, F$ are the vertices of a rhombus with a... | 176. 177) Since $\angle D C B=\angle C D B$ and $B F \perp C D$ (Fig. 18), the height $B M$ (where $M$ is the intersection point of $C D$ and $B F$) will also be the median of the isosceles triangle $C B D$, i.e., $C M=M D$. Then, the equality of $M E$ and $M F$ is established as the projections of the inclined lines $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,752 |
177. Tangents $A T_{1}$ and $A T_{2}$ are drawn to a circle, forming an angle of $60^{\circ}$. Through a point $M$, taken arbitrarily on the circle, lines $T_{1} M$ and $T_{2} M$ are drawn, intersecting the tangents $A T_{2}$ and $A T_{1}$ at points $Q_{1}$ and $Q_{2}$, respectively. Prove that the segments $T_{1} Q_{1... | 177. Establish the equality of triangles $A T_{1} Q_{1}$ and $T_{2} T_{1} Q_{2}$, using the property of angles related to the circle and the fact that triangle $A T_{1} T_{2}$ is equilateral. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,753 |
179. Inside an acute-angled triangle $A B C$, there is a point $M$ such that $\angle M A B - \angle M B A = \angle M B C - \angle M C B = \angle M C A - \angle M A C$. Prove that the point $M$ is the center of the circumcircle of triangle $A B C$. | 179. Assume the opposite, i.e., $\angle M A B > \angle M B A$. Then $\angle M B C > \angle M C B$ and $\angle M C A > \angle M A C$. By considering the sides $A M, M B$, and $M C$ of the formed triangles, obtain a contradiction: $A M < B M < M C < A M$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,755 |
180. A diameter of the circumscribed circle around a triangle is drawn through one of its vertices. Prove that the bisector of the angle between the diameter and the altitude dropped from the same vertex to the opposite side is also the bisector of the angle of the triangle at that vertex. | 180. Extend the height until it intersects with the circle and connect the obtained point with the second end of the diameter. Then establish that the drawn segment is parallel to the side of the triangle, and, therefore, the bisector divides the arc subtended by this side in half. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,756 |
181. Prove that the point symmetric to the center of the circle relative to the midpoint of the midline of a triangle inscribed in this circle lies on the height perpendicular to this midline. | 181. Establish that if point $M$ is the midpoint of the midline $A_{1} B_{1}$ of triangle $A B C$ and $O N$ is the perpendicular dropped from the center $O$ of the circumscribed circle to side $A B$, then $N C$ will intersect $A_{1} B_{1}$ at point $M$ and $C M = M N$. Consequently, the segment $N O$ symmetrical to poi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,757 |
182. A line drawn through the vertex of a triangle and the midpoint of the opposite side passes through the center of the circle circumscribed around this triangle. Prove that if the triangle is scalene, then it is a right triangle. | 182. Assume the opposite; then the median $B M_{2}$ of triangle $A B C$ bisects side $A C$, and also bisects it perpendicularly from the center $O$ of the circle dropped to this side. The contradiction rejects the assumption. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,758 |
183. Points $C_{1}$ and $C_{2}$ are obtained by reflecting vertex $C$ of triangle $A B C$ across the angle bisectors of angles $A$ and $B$. Prove that the point of tangency of the incircle of triangle $A B C$ divides the segment $C_{1} C_{2}$ in half and that this segment is seen from the center of the circle at an ang... | 183. Consider the triangle $C C_{1} C_{2}$ and establish that the point $O^{\prime}$, the intersection of the angle bisectors of angles $A$ and $B$, is equidistant from the points $C$, $C_{1}$, and $C_{2}$. From this, it follows that the perpendicular $O^{\prime} K$, dropped from point $O^{\prime}$ to side $A B$, bisec... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,759 |
184. On the segment $C H$ of triangle $A B C$ (where $H$ is the orthocenter), a circle is constructed with $C H$ as its diameter, intersecting sides $A C$ and $B C$ at points $M$ and $N$ respectively. Prove that the tangents to the circle at points $M$ and $N$ intersect at the midpoint of side $A B$. | 184. Establish the equality of angles $NAB$ and $ANK$ ($K$ - the intersection point of $AB$ with the tangent at point $N$), using the properties of angles with perpendicular sides, as well as the equality of angles related to the circle. Then consider the right triangle $ANB$ and show the equality of angles $KNB$ and $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,760 |
185. A right triangle is divided by the height dropped from the vertex of the right angle to the hypotenuse into two triangles, into which circles are inscribed. Prove that the line of centers of these circles is perpendicular to the bisector of the right angle of the given triangle. | 185. Consider the triangle $\mathrm{CO}_{1} \mathrm{O}_{2}$ with vertices at the centers $O_{1}$ and $O_{2}$ of the inscribed circles and at the vertex $C$ of the right angle of the given triangle $A B C$ and establish that the segments of the angle bisectors of angles $A$ and $B$ are altitudes of the triangle $\mathrm... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,761 |
186. A right triangle $A B C$ with height $\mathrm{CH}_{3}$, dropped to the hypotenuse, is divided into two triangles $A \mathrm{CH}_{3}$ and $B \mathrm{CH}_{3}$. Prove that the centers $O, O_{1}, O_{2}$ of the circles inscribed in triangles $A B C, A C H_{3}, H_{3} B C$ are equidistant from the point of tangency of th... | 186. Draw the bisectors $C C_{1}$ and $C C_{2}$ of angles $A C H_{3}$ and $B C H_{3}$; establish that triangle $C C_{1} O_{2}$ is right-angled and isosceles, using the perpendicularity of $B O^{\prime}$ and $C C_{1}$. Show further that angle $C_{1} O^{\prime} C_{2}$ is a right angle, as a central angle of the circle ci... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,762 |
189. The bisectors of the angles formed by the opposite sides of a convex quadrilateral are perpendicular. Prove that a circle can be circumscribed around such a quadrilateral. | 189. Consider the quadrilaterals into which the given quadrilateral is divided by its bisectors. Establish that they pairwise have two respectively equal angles, and therefore the sums of the other two angles are also equal. From these equalities, it follows that the sums of the opposite angles of the given quadrilater... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,765 |
190. Three lines pass through one point $S$, forming six angles of $60^{\circ}$. Prove that the sum of the distances from any point to two of the lines is equal to the distance to the third line. | 190. Through the given point $P$, draw a line parallel to that one of the three given lines from which the point $P$ is most distant. Consider the resulting equilateral triangle and establish that the sum of the distances from point $P$ to two of its sides is equal to the height of the triangle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,766 |
191. If a quadrilateral has two of the three properties listed below, namely: 1) the diagonals of the quadrilateral are perpendicular, 2) the quadrilateral is cyclic, 3) the perpendicular dropped from the vertex of one side to the opposite side passes through the point of intersection of the diagonals, then it has the ... | 191. Let, for example, the first two conditions be satisfied, then draw a line through the midpoint of one of the sides and the point of intersection of the diagonals, and then, using the properties of inscribed angles and angles with correspondingly perpendicular sides, prove that this line is perpendicular to the opp... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,767 |
192. Through the vertex $C$ of a right angle, two arbitrary lines are drawn. From points $A$ and $B$, taken on the sides of the given angle, perpendiculars $A A_{1}$ and $A A_{2}, B B_{1}$ and $B B_{2}$ are dropped to these lines ($A_{1}, A_{2}, B_{1}, B_{2}$ - the feet of the perpendiculars). Prove that the lines $A_{... | 192. Describe circles around quadrilaterals $A A_{1} C A_{2}$ and $B B_{1} C B_{2}$ and use the properties of inscribed angles and angles with respectively perpendicular sides. Prove that the sides of angles $A A_{2} A_{1}$ and $B_{1} B_{2} C$ are respectively perpendicular. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,768 |
193. From vertices $A$ and $B$ of triangle $ABC$, heights $AH_1$ and $BH_2$ are dropped; from points $H_1$ and $H_2$, perpendiculars $H_1A_1$ and $H_2B_1$ are drawn to the same sides $AC$ and $BC$. Prove that line $A_1B_1$ is parallel to line $AB$. | 193. Consider two inscribed quadrilaterals $A B H_{1} H_{2}$ and $A_{1} B_{1} H_{1} H_{2}$ and establish the parallelism of lines $A B$ and $A_{1} B_{1}$ based on the equality of corresponding angles. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,769 |
194. Prove that if the bases of the perpendiculars dropped from a point $M$ in the plane of triangle $A B C$ to its sides lie on one straight line, then the point $M$ lies on the circumcircle of triangle $A B C$. Verify the validity of this theorem for the case when lines are drawn through point $M$ at equal angles to ... | 194. Using the properties of three inscribed quadrilaterals and the equality of vertical angles formed by the line connecting the bases of the perpendiculars and side $B C$ of the triangle, establish that angle $C M B$ is equal to angle $A$ or supplements it to $180^{\circ}$ (the converse of Simson's theorem). | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,770 |
195. Through the vertex $C$ of triangle $A B C$, a secant is drawn, from vertices $A$ and $B$ perpendiculars $A A_{1}$ and $B B_{1}$ are dropped to this secant. From points $A_{1}$ and $B_{1}$, perpendiculars $A_{1} A_{2}$ and $B_{1} B_{2}$ are dropped to sides $B C$ and $A C$, respectively, intersecting at point $D$. ... | 195. Establish that the quadrilateral $A_{1} C_{1} B_{1} D$ is cyclic, considering the cyclic quadrilaterals $A A_{1} C H_{3}, B B_{1} C H_{3}$, $A_{2} C B_{2} D$. Then prove that the points $D, C, H_{3}$ are collinear. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,771 |
197. Prove that if the midline of a quadrilateral, corresponding to one pair of opposite sides, forms equal angles with the other two sides, then these sides are either equal or parallel. | 197. Parallelly translate sides $A B$ and $C D$ so that their ends $B$ and $C$ coincide at the midpoint $M_{2}$ of side $B C$. In the resulting triangle $A^{\prime} M_{2} D^{\prime}$, the median $M_{2} M_{4}$ is also a bisector. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,773 |
199. A quadrilateral is inscribed in a circle, the diagonals of which are perpendicular. The bases of the perpendiculars dropped from the intersection point of the diagonals of this quadrilateral to its sides are the vertices of a second quadrilateral. Prove that a circle can be inscribed in the new quadrilateral and a... | 199. Establish that the intersection point of the diagonals of the given quadrilateral is the intersection point of the angle bisectors of the new quadrilateral. Then establish that the sum of the opposite angles of this quadrilateral is expressed through the angles of the given quadrilateral $ABCD$ as follows:
$4 d-2... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,775 |
200. A quadrilateral inscribed in a circle is divided by one diagonal into two triangles, in which circles are inscribed, and by the second diagonal - into two other triangles, in which circles are also inscribed. Prove that the centers of these circles are the vertices of a rectangle. | 200. Prove that if $M, N, E, F$ are the midpoints of the arcs $AB, CD, BC$, and $AD$, into which the vertices of the quadrilateral $ABCD$ divide the circle (Fig. 19), then the centers $O_{1}$ and $O_{2}, O_{3}$ and $O_{4}$ of the circles inscribed in the given triangles lie at the points of intersection of the bisector... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,776 |
201. Perpendiculars are dropped from the foot of each altitude of a triangle to the other two sides of the triangle. Prove that the bases of all six perpendiculars lie on a single circle. | 201. If the bases of the perpendiculars are denoted respectively as $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}$, then establish that the following sets of four points lie on the same circle: $A_{1}, A_{2}, C_{1}, B_{2}$; $A_{1}, A_{2}, B_{1}, C_{2}$; $A_{1}, A_{2}, C_{1}, C_{2}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,777 |
202. Prove that if the orthocenter of a triangle lies on the circle passing through the midpoints of its sides, then the triangle is a right triangle. | 202. Prove by contradiction, using the property of the Euler circle for a triangle.
Fig. 19
 is rotated about its center by an angle less than $120^{\circ}\left(\frac{1}{n} \cdot 360^{\circ}\right)$. Prove that the sides of the given triangle ($n$-gon) and the sides of the rotated triangle ($n$-gon) are the sides of a regular hexagon ($2n$-gon), into which a circle can be insc... | 203. If triangle $A_{1} A_{2} A_{3}$ occupies after rotation the position $B_{1} B_{2} B_{3}$, then establish the parallelism of the lines $A_{2} B_{2}, A_{3} B_{1}$ and $A_{1} B_{3}$ and the symmetry of the resulting hexagon relative to the diameter perpendicular to $A_{2} B_{2}$. Similarly, discover the presence of t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,779 |
205. On the hypotenuse $AB$ of a right triangle $ABC$, an arbitrary point $P$ is given. Prove that the circles circumscribed around triangles $APC$ and $BPC$ are orthogonal.
## GL $L$ in $A \quad I I I$
## PARALLELISM AND CONTINUITY. SIMILARITY
## § 11. Proportional segments (affine problems) | 205. If $\angle A C O_{1}=\alpha, \angle B C O_{2}=\beta$ (where $O_{1}$ and $O_{2}$ are the centers of the circumcircles of triangles $A C P$ and $B C P$), then $\angle A P C=90^{\circ}-\alpha, \angle B P C=90^{\circ}+\beta$ (fig. 20). Therefore, $\left(90^{\circ}-\alpha\right)+\left(90^{\circ}+\right.$ $+\beta)=180^{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,781 |
206. Point $C$ divides the segment $A B$ internally in the ratio $k$, and point $C_{1}$ divides the same segment externally in the same ratio. Prove that the midpoint of segment $C C_{1}$ divides the segment $A B$ externally in the ratio $k_{2}$. | 206. It is not difficult to establish that $A M=\frac{1}{2}\left(A C+A C_{1}\right)$ and $B M=$ $=\frac{1}{2}\left(B C+B C_{1}\right)$. But $A C=\frac{k}{1+k} A B, C B=\frac{A B}{1+k} ; A C_{1}=-\frac{k}{1-k} \cdot A B$; $B C_{1}=\frac{A B}{1-k}$. From this, we have: $A M: B M=-k^{2}$, i.e., the division of segment $A ... | k^2 | Geometry | proof | Yes | Yes | olympiads | false | 43,782 |
207. On lines $m$ and $n$, points $A, B, C$ and $A_{1}, B_{1}, C_{1}$ are taken respectively such that the lines $A A_{1}, B B_{1}, C C_{1}$ are parallel. Prove that the points $A_{2}, B_{2}, C_{2}$, which divide the segments $A A_{1}, B B_{1}, C C_{1}$ in equal ratios, lie on a line that, together with lines $m$ and $... | 207. Use the theorem about the intersection of a pencil of lines by parallel lines and show that a secant drawn through point $A_{2}$ and the intersection point of lines $m$ and $n$ will divide the segments $B B_{1}$ and $C C_{1}$ in the ratio $A A_{2}: A_{2} A_{1}$, i.e., this secant will pass through points $B_{2}$ a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,783 |
208. Through the vertex $C$ of triangle $A B C$, a secant is drawn, intersecting the midline $M_{1} M_{3}$ and the extension of the midline $M_{2} M_{3}$ at points $P$ and $Q$, respectively. Prove that $A Q$ and $B P$ are parallel. | 208. Extend $M_{1} M_{3}$ to intersect $A Q$ at point $S$ and, using the main theorem of proportional segments, prove that $P M_{3}=M_{3} S$, i.e., quadrilateral $A P B S$ is a parallelogram and lines $A Q$ and $B P$ are parallel. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,784 |
209. Through the vertices of triangle $A B C$ and a point $P$ taken inside it, lines are drawn, intersecting the opposite sides at points $A_{1}, B_{1}$, and $C_{1}$, respectively. The line drawn through point $C_{1}$ parallel to $A A_{1}$ intersects sides $A C$ and $B C$ of the triangle at points $M$ and $N$, respecti... | 209. Consider the pencils of lines with vertices at points $C$ and $B$, intersected by the parallel lines $A A_{1}$ and $M N$, and establish that $M N: C_{1} N=A A_{1}: P A_{1}=C_{1} N: N D$, from which the required equality follows. | NC_{1}^{2}=MN\cdotND | Geometry | proof | Yes | Yes | olympiads | false | 43,785 |
210. On the side of a triangle, two equal segments are laid out from the corresponding vertices of this side. Through the obtained points, lines parallel to the corresponding sides of the triangle are drawn. Prove that the constructed lines intersect on the median of the triangle drawn to the initially chosen side. | 210. Let the lines drawn through points $E$ and $F$ of the base $AB$ of triangle $ABC$ intersect at point $P$. Then the line $CP$ will intersect side $AB$ at such a point $S$ that $AE: ES = CP: PS = BF: FS$. But by the condition $AE = BF$, so $ES = FS$, i.e., $CS$ is the median of triangle $ABC$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,786 |
211. Through a point $P$ located inside the triangle $ABC$, three rays are drawn: ray $p_{1}$, parallel to $AB$, intersecting $BC$ at point $A_{1}$, ray $p_{2}$, parallel to $BC$, intersecting $CA$ at point $B_{1}$, ray $p_{3}$, parallel to $CA$, intersecting $AB$ at point $C_{1}$. Prove that the following relationship... | 211. All points located inside the triangle or on its sides satisfy this condition, and therefore they belong to the geometric locus. If a point lies outside the triangle, then the equality given in the condition does not hold. To prove this, draw cevians through point $P$ and consider the sum of the ratios of the segm... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,787 |
212. Prove that if a segment enclosed between the opposite sides of a quadrilateral and drawn through the point of intersection of its diagonals parallel to one of the two other sides is bisected by this point, then the quadrilateral is a trapezoid or a parallelogram. | 212. Consider two pairs of similar triangles formed and establish that the diagonals of the given quadrilateral are divided by their point of intersection in equal ratios. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,788 |
214. Through point $M$ of side $A B$ of triangle $A B C$, lines $M N$ and $M K$ are drawn parallel to $A C$ and $B C$ respectively. An arbitrary secant passing through vertex $C$ intersects segment $M N$ at point $P$, and the extension of $K M$ at point $Q$. Prove that line $A Q$ is parallel to $B P$. | 214. Establish that $A M: M B=A K: K C=M M_{1}: M P$ (where $M_{1}$ is the intersection point of lines $A Q$ and $M N$), from which it follows that triangles $A M M_{1}$ and $B M P$ are similar, and therefore, lines $A Q$ and $B P$ are parallel. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,790 |
215. Through two opposite vertices of a parallelogram, lines are drawn, each of which meets the other two sides at points $M$ and $N, K$ and $L$. Prove that the resulting points are vertices of a trapezoid or a parallelogram. Formulate and prove the converse theorem. | 215. If lines $M N$ and $L K$ are drawn through vertices $A$ and $C$ respectively, then establish that triangles $M B L$ and $N D K$ are similar. For this, consider a series of similar triangles and show that $M B: B L=K D: D N$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,791 |
217. An arbitrary point $P$ on the lateral side $AB$ of trapezoid $ABCD$ is connected to vertices $C$ and $D$. Through the vertices $A$ and $B$ of the trapezoid, lines are drawn parallel to segments $PC$ and $PD$, respectively. Prove that the constructed lines intersect at a point on side $CD$. Determine the validity o... | 217. Assume the opposite, then these lines will intersect side $C D$ at points $Q_{1}$ and $Q_{2}$, respectively. Then extend $A Q_{1}$ to intersect $B C$ at point $K$ and $B Q_{2}$ to intersect $A D$ at point $L$ and establish that $\frac{B C}{C K}=\frac{B P}{P A}=\frac{D L}{A D}$, and therefore, $\frac{D Q_{1}}{Q_{1}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,793 |
218. Prove that if the center of homothety of two homothetic triangles is the centroid of one of the triangles, then it is also the centroid of the other. | 218. If two triangles are homothetic, then their centroids are also homothetic. But if one centroid coincides with the center of homothety, then the homothetic point to it also coincides with the center of homothety, for the center of homothety is an invariant point of the transformation. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,794 |
219. The midpoints of two opposite sides of a quadrilateral are connected to a point located on the diagonal. Prove that the two constructed lines divide the other two sides of the quadrilateral in equal ratios. | 219. Let a line passing through point $P$, lying on the extension of diagonal $D B$, and the midpoint $M_{1}$ of side $A B$, intersect side $A D$ at point $K$. If perpendiculars $A A_{1}, B B_{1}, D D_{1}$ are dropped from this line and the similar triangles $A A_{1} K$ and $D D_{1} K, P D D_{1}$ and $P B B_{1}$ are co... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,795 |
220. On the diagonal $AC$ of trapezoid $ABCD$ (side $AD$ is parallel to $BC$, $AD > BC$), a segment $AP$ is laid off equal to $OC$, and on the diagonal $DB$ - a segment $DQ$ equal to $OB$ ($O$ is the point of intersection of the diagonals). Prove that the lines drawn through points $B$ and $P$, $C$ and $Q$, cut off equ... | 220. Segments that enter the obvious proportion $\frac{A O}{O C}=\frac{D O}{O B}$, replace with equal segments and from the resulting proportion $\frac{A P}{P C}=\frac{D Q}{Q B}$ and similar triangles $A M P$ and $B P C, D N Q$ and $B Q C$ establish the equality of segments $A M$ and $D N$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,796 |
221. A line drawn through vertex $C$ of trapezoid $A B C D$ (side $A D$ is parallel to side $B C$) parallel to side $A B$, cuts off a segment $D P$ on diagonal $D B$, equal to segment $O B$ (where $O$ is the point of intersection of the diagonals of the trapezoid). Prove that $A D^{2}=$ $=B C^{2}+A D \cdot B C$. | 221. From the similarity of triangles $A O D$ and $B O C$, $A B D$ and $K P D$ (where $K$ is the intersection point of lines $C P$ and $A D$), considering the equality $B O=P D$, it follows that $B C: A D=B O: O D=P D: B P=$ $=K D: A K=(A D-B C): B C$.
From this, it follows that $A D^{2}=B C^{2}+$ $+A D \cdot B C$. | AD^{2}=BC^{2}+AD\cdotBC | Geometry | proof | Yes | Yes | olympiads | false | 43,797 |
222. Through the vertex $C$ of the smaller base $B C$ of the trapezoid $A B C D$, a line parallel to the side $A B$ is drawn and intersects the diagonal $B D$ at point $P$. Prove that if $D P=B O$ (where $O$ is the point of intersection of the diagonals), then a line parallel to $C D$ and passing through the vertex $B$... | 222. Establish the equality of ratios: $C O: A C=(A D-B C): A D$ (see the hint for problem № 221) and $A Q: A C=A L: A D$ (from the similarity of triangles $A C D$ and $A Q L$), where $L$ is the point of intersection of lines $A D$ and $B Q$. But $A D-B C=A L$, therefore, $C O=A Q$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,798 |
224. The sides of triangle $ABC$ are divided in the same ratio $\lambda$ in the direction of traversal along its contour. The sides of the resulting triangle $A_{1} B_{1} C_{1}$ are also divided in the same ratio $\lambda$ in the opposite direction of traversal. Prove that the third resulting triangle $A_{2} B_{2} C_{2... | 224. Draw the line $C_{1} M$ parallel to $A C$ and intersecting $B C$ at point $M$ and $A_{1} B_{1}$ at point $N$, and establish that $B_{1} N: B_{1} C_{2}=\lambda+1$ and $B_{1} C_{1}: B_{1} A_{2}=\lambda+1$, i.e., $A_{2} C_{2}$ is parallel to $C_{1} M$ and side $A C$. Similarly, it can be proven that $C_{2} B_{2} \| C... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,800 |
225. Prove that if a line passing through the midpoints of opposite sides of a quadrilateral passes through the point of intersection of its diagonals, then the quadrilateral is a trapezoid or a parallelogram. | 225. Extend the segments $O M_{2}$ and $O M_{4}$ of the midline $M_{2} M_{4}$ ( $O$ - the point of intersection of the diagonals $A C$ and $B D$ ) beyond the points $M_{2}$ and $M_{4}$. On these extensions, lay off the segments $O M_{2}=M_{2} K$ and $O M_{4}=$ $=M_{4} L$ and consider the parallelograms $A O D L$ and $B... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,801 |
226. A trapezoid is inscribed in a quadrilateral, the parallel sides of which are parallel to its diagonal. Prove that the non-parallel sides of the trapezoid intersect on the other diagonal of the quadrilateral. | 226. Establish that the non-parallel sides $M N$ and $L K$ of trapezoid $M N K L$ divide the diagonal $D B$ externally in the same ratio. For this, apply Menelaus' theorem to triangle $A B D$ and the transversal $M N$, as well as to triangle $B C D$ and the transversal $L K$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,802 |
227. Two lines $a$ and $b$ are intersected by two other lines at points $A_{1}$ and $A_{2}, B_{1}$ and $B_{2}$. On the secant $A_{1} B_{1}$, a point $P$ is given, which is connected to points $A_{2}$ and $B_{2}$. Through the midpoint $M_{1}$ of segment $A_{1} B_{1}$, lines $a_{1}$ and $b_{1}$ are drawn parallel to line... | 227. Obviously, $\frac{A_{2} X}{X P}=\frac{A_{1} M_{1}}{M_{1} P}, \frac{P M_{1}}{M_{1} B_{1}}=\frac{P Y}{Y B_{2}}$. From this, by multiplying term by term, we get: $\frac{A_{2} X}{X P} \cdot \frac{P Y}{Y B_{2}}=-1$. But according to Menelaus' theorem, $\frac{A_{2} X}{X P} \cdot \frac{P Y}{Y B_{2}} \cdot \frac{B_{2} M^{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,803 |
228. A straight line meets the sides $A B, B C, C D, D A$ of a quadrilateral $A B C D$ or their extensions at points $P, Q$, $R, S$ respectively. Prove that the segments formed satisfy the equation:
$$
\frac{\overline{A P}}{\overline{P B}} \cdot \frac{\overline{B Q}}{\overline{Q C}} \cdot \frac{\overline{C R}}{\overli... | 228. Apply Menelaus' theorem to the triangles into which the given quadrilateral is divided by one of its diagonals and the transversal $P S$; multiply the obtained equalities. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,804 |
230. Four lines belonging to the same pencil are intersected by two lines at points $A, B, C, D$ and $A_{1}, B_{1}, C_{1}, D_{1}$ respectively. Prove that the segments formed on the intersecting lines satisfy the equation:
$$
\frac{\overline{A C}}{\overline{C B}}: \frac{\overline{A D}}{\overline{D B}}=\frac{\overline{... | 230. Using the result of problem No. 229, consider the quadrilateral $A B B_{1} A_{1}$ and the secant $C C_{1}$, and then the same quadrilateral and the secant $D D_{1}$. Equate the obtained equalities. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,806 |
231. A secant is drawn through the point $M$ of intersection of the diagonals of a quadrilateral. The segment of this secant, enclosed between one pair of opposite sides of the quadrilateral, is bisected by the point $M$. Prove that the segment of the secant, enclosed between the extensions of the other pair of opposit... | 231. Let the secant $E M F$ meet
Fig. 22 the sides $A D$ and $B C$ of the quadrilateral $A B C D$ at points $E$ and $F$, and the extensions of sides $A B$ and $C D$ at points $K$ and $L$. ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,807 |
232. On the sides $AC$ and $BC$ of triangle $ABC$, there are two points $M_{1}$ and $M_{2}, N_{1}$ and $N_{2}$ such that $A M_{i}: M_{i} C = C N_{i}: N_{i} B = k_{i} (i=1,2)$. Prove that the point $P$, where the segments $M_{1} N_{1}$ and $M_{2} N_{2}$ intersect, divides each of these segments in the ratio $k_{2}$ and ... | 232. Using the properties of proportions, establish the equality of the following ratios: $\frac{A M_{1}}{C N_{1}}=\frac{A M_{2}}{C N_{2}}=\frac{M_{1} M_{2}}{N_{1} N_{2}}$ (Fig. 22). Further applying Menelaus' theorem to triangle $M_{1} N_{1} C$ and the transversal $M_{2} N_{2}$, as well as to triangle $M_{2} N_{2} C$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,808 |
233. The sides of a triangle are divided in the same ratio when traversing its contour in one direction. Prove that the points of division are the vertices of a triangle whose centroid coincides with the centroid of the given triangle. Check the validity of this theorem for quadrilaterals. | 233. Draw the midline $A_{2} C_{2}$ of triangle $ABC$, which intersects side $A_{1} C_{1}$ of triangle $A_{1} B_{1} C_{1}$ at point $P$. Based on the result of problem № 232, we conclude that $B_{1} P$ is the median of triangle $A_{1} B_{1} C_{1}$, intersecting the median $C C_{2}$ of triangle $ABC$ at point $M$ such t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,809 |
234. Through the point $M$ of intersection of the diagonals $A C$ and $B D$ of the quadrilateral $A B C D$, a secant is drawn parallel to the side $A B$ and intersects the other three sides or their extensions at points $P, Q, R$ (point $R$ lies on the line $C D$). Prove that the segments formed on the secant satisfy t... | 234. The lines emanating from points $C$ and $D$ intercept proportional segments on lines $Q R$ and $A B$: $M P: P R=A B: B S$ and $Q M: M R=A B: B S$, i.e., $M P: P R=Q M: M R$, or $\frac{M P+P R}{P R}=$ $=\frac{Q M+M R}{M R}$. From this, $R M^{2}=R P \cdot R Q$. | RM^{2}=RP\cdotRQ | Geometry | proof | Yes | Yes | olympiads | false | 43,810 |
236. On a line, there are two pairs of points $A_{1}$ and $A_{2}, B_{1}$ and $B_{2}$. Prove that there exists a unique point $P$ on the line such that the equality $P A_{1} \cdot P A_{2}=P B_{1} \cdot P B_{2}$ holds. | 236. Draw a circle through points $A_{1}$ and $A_{2}$, intersecting the circle drawn through points $B_{1}$ and $B_{2}$. If the segments $A_{1} A_{2}$ and $B_{1} B_{2}$ do not have a common midpoint, then both circles intersect at points located on a line intersecting the given line at the desired point. Uniqueness is ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,812 |
238. On the median $C M_{3}$ of triangle $A B C$, a point $M$ is given such that $\frac{C M}{M M_{3}}=\alpha$. Through the point $M$, an arbitrary secant is drawn, intersecting the sides $A B, B C, C A$ of the triangle or their extensions at points $P, Q, R$, respectively, with points $Q$ and $R$ lying on the same side... | 238. Let the ratio $\frac{A M_{3}}{M_{3} P}$ be denoted by $\beta$ (Fig. 23). Then, applying Menelaus' theorem to triangle $B C M_{3}$ and the transversal $P R$, we have: $\quad \frac{Q C}{C B}=\frac{\alpha}{\beta-\alpha-1}$. Based on the same theorem, we determine: $\frac{P M}{M Q}=\frac{\beta-\alpha-1}{\alpha \beta}$... | \frac{1}{MQ}+\frac{1}{MR}=\frac{2}{\alpha}\cdot\frac{1}{MP} | Geometry | proof | Yes | Yes | olympiads | false | 43,814 |
239. Opposite vertices of one parallelogram are located respectively on the opposite sides (or their extensions) of another parallelogram. Prove that both parallelograms have a common center of symmetry. | 239. Establish that the segments of each diagonal of the given parallelogram, lying outside the inscribed parallelogram, are equal. From this, it follows that the point of intersection of the diagonals of the given parallelogram is the center of symmetry of the inscribed parallelogram.
, and show that if point $F$ on side $AB$ moves to point $F_{1}$ on segment $DP$, then $EF_{1}$ is parallel to $CP$. In triangle $CDP$, divide side $CP$ at point $L$ in the ratio $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,824 |
250. On the side $AB$ of triangle $ABC$, a point $P$ is given, through which lines parallel to its medians $AM_{1}$ and $BM_{2}$ are drawn and intersect the corresponding sides of the triangle at points $A_{1}$ and $B_{1}$. Prove that the midpoint of segment $A_{1}B_{1}$, point $P$, and the centroid $G$ of the given tr... | 250. Establish that each of the medians divides the segments $P A_{1}$ and $P B_{1}$ in the ratio $2: 1$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,826 |
251. In the plane of triangle $A B C$, a point $P$ is given. Through the centroids $G_{3}, G_{1}, G_{2}$ of triangles $A B P, B C P, C A P$, lines are drawn, respectively parallel to the lines $C P, A P, B P$. Prove that these lines intersect at the centroid of the given triangle. | 251. Notice that triangle $G_{1} G_{2} G_{3}$ is homothetic to the given one with a homothety coefficient equal to $\frac{1}{3}$. Due to homothety, the lines drawn through points $G_{1}, G_{2}, G_{3}$ parallel to $A P$, $B P, C P$ respectively, intersect at one point located on the median of the given triangle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,827 |
252. In the plane of triangle $ABC$, a point $P$ is taken, through which lines parallel to the medians of the triangle are drawn and intersect the corresponding sides at points $A_{1}, B_{1}, C_{1}$. Prove that the centroid of triangle $A_{1} B_{1} C_{1}$ lies at the midpoint $M$ of the segment connecting point $P$ and... | 252. Apply the solution of problem 251 to the triangle $A_{1} B_{1} C_{1}$ and point $M$. Also, use the result of problem 250 to establish that the centroid of triangle $M A_{1} B_{1}$ lies on the midline of the parallel lines $C M_{3}$ and $P C_{1}$, by drawing a line through point $M$ parallel to $A B$ (fig. 26). | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,828 |
253. Two triangles $A_{1} B_{1} C_{1}$ and $A_{2} B_{2} C_{2}$ have a common centroid. Prove that if the lines $A_{1} A_{2}, B_{1} B_{2}, C_{1} C_{2}$ are parallel, then the corresponding sides of the triangles intersect at points which, together with the common centroid, lie on the same line. Prove the converse theore... | 253. If the lines connecting the corresponding vertices of two triangles are parallel, then these triangles have a perspective axis, on which the corresponding sides of these triangles intersect (Desargues' theorem). One triangle is obtained from the other by a skew transformation to the perspective axis. However, in t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,829 |
255. The lines $A B, A C, A D$ are intersected by a line at points $P, Q, R$ respectively. Prove that if points $A, B, C, D$ are vertices of a parallelogram (with $A C$ as a diagonal), then the following relationship holds between the formed segments:
$$
\frac{A B}{A P}+\frac{A D}{A R}=\frac{A C}{A Q}
$$
(consider th... | 255. Draw lines through points $D$ and $C$ parallel to the secant and intersecting line $A B$ at points $N$ and $M$. Use the proportions $A D: A R=A N: A P, A C: A Q=A M: A P$ and the equality of segments $A N$ and $B M$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,831 |
256. On the sides $A B, B C, C D, D E$ of the pentagon $A B C D E$, points $P, Q, R, S$ are given respectively such that $A P: P B = D R: R C = k_{1}$, $B Q: Q C = E S: S D = k_{2}$. On the segments $P R$ and $Q S$, points $M$ and $N$ are constructed respectively such that $P M: M R = k_{2}$, $S N: N Q = k_{1}$. Prove ... | 256. Establish that $R N$ meets $B E$ at point $L$ such that $E L: L B=k_{1}$. Therefore, $P L \| A E$, and $P L=\frac{A E}{k_{1}+1}$. Similarly, we find that $L N: N R=k_{2}$ and $P M: M R=k_{2}$. Therefore, $L P \| N M$, and $M N=\frac{P L}{k_{2}+1}$. Thus, $M N \| A E$ and $M N=$ $=\frac{A E}{\left(k_{1}+1\right)\le... | MN=\frac{AE}{(k_{1}+1)(k_{2}+1)} | Geometry | proof | Yes | Yes | olympiads | false | 43,832 |
257. On a straight line, there are three equal segments $A B, B C, C D$. Through the point $S$, not lying on the given line, the lines $S A$, $S B$, $S C$, $S D$ are drawn. An arbitrary secant, not passing through the point $S$, intersects these lines at points $A_{1}, B_{1}, C_{1}$, $D_{1}$ respectively. Prove that
$... | 257. Through points $A, B, C, D, S$ draw parallel lines intersecting the secant at points $A_{0}, B_{0}, C_{0}, D_{0}, S_{0}$ (Fig. 28). Obviously, $\frac{A A_{1}}{A_{1} S}+\frac{D D_{1}}{D_{1} S}=\frac{A A_{0}+D D_{0}}{S S_{0}}$ and $\frac{B B_{1}}{B_{1} S}+\frac{C C_{1}}{C_{1} S}=$ $=\frac{B B_{0}+C C_{0}}{S S_{0}}$.... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,833 |
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