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84. Is the following statement true: if in two quadrilaterals $A B C D$ and $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ the corresponding sides are equal ( $A B=A^{\prime} B^{\prime}, B C=B^{\prime} C^{\prime}$, etc.), then the quadrilaterals are equal? | $\triangleright$ Recalling the previous problem, we can formulate the question as follows: will a hinged quadrilateral be rigid? It is immediately clear that no - by taking the opposite vertices in your hands, you can move them closer and further apart. The diagram shows two positions of the hinges - two different quad... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,905 |
85. Provide an example of triangles $A B C$ and $A_{1} B_{1} C_{1}$, in which $A B=A_{1} B_{1}, A C=A_{1} C_{1}$, $\angle A B C=\angle A_{1} B_{1} C_{1}$, but which are nevertheless not equal.
This problem shows that in the criterion for the equality of triangles "by two sides and the angle between them," the words "b... | $\triangleright$ Let's draw an acute angle with vertex $B$. On one side, take any point $A$. On the other side, using a compass, find two points $C$ and $C_{1}$ at equal distances from $A$. Then triangles $A B C$ and $A B C_{1}$ will be the desired ones. $\triangleleft$
 and a bisector (divides the angle in half). | $\triangleright$ Let's denote our triangle as $A B C$ (base $B C$, lateral sides $A B$ and $A C$). The median $A M$ connects vertex $A$ to the midpoint of the base $B C$. It divides the triangle into two halves $A B M$ and $A C M$. It is easy to see that these halves are equal by the third criterion (one side is common... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,907 |
90. In triangle $ABC$, the median $AM$, connecting point $A$ to the midpoint $M$ of side $BC$, is an altitude, that is, perpendicular to side $BC$. Prove that triangle $ABC$ is isosceles: sides $AB$ and $AC$ are equal. | $\triangle$ Consider triangles $A B M$ and $A C M$. In these triangles, the angles at vertex $M$ are right angles by the given condition, $B M=C M$, and side $A M$ is common. Therefore, by the first criterion of triangle congruence, they are congruent. In particular, the sides opposite the right angle are equal: $A B=A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,908 |
91. In triangle $A B C$, the median $A M$, connecting point $A$ to the midpoint $M$ of side $B C$, is the bisector of angle $A$. Prove that triangle $A B C$ is isosceles: sides $A B$ and $A C$ are equal.
.
Thus, this problem is more complicated tha... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,909 |
92. Prove that any point on the perpendicular bisector of segment $A B$ is equidistant (located at the same distance) from points $A$ and $B$. | $>$ Let $M$ be the midpoint of segment $AB$, and $X$ lies on the perpendicular bisector. Connect $X$ with points $A$ and $B$. In triangles $AMX$ and $BMX$, two angles are right, one side ($MX$) is common, and $AM = MB$. Therefore, these triangles are congruent by the first criterion, in particular, $AX = XB. \trianglel... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,910 |
93. Prove that any point equidistant from points $A$ and $B$ lies on the perpendicular bisector of the segment $A B$. | $\triangleright$ Let point $X$ be equidistant from points $A$ and $B$. Connect it to these points, and we get an isosceles triangle $A X B$. Connect point $X$ to the midpoint of side $A B$, which we will denote as $M$. We get the median $X M$ of this isosceles triangle, which, as we know, is also an altitude. Therefore... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,911 |
94. An isosceles triangle is divided into two by a segment connecting the vertex to a certain point on the base. Prove that in these triangles, two pairs of equal sides and one pair of equal angles can be found. | $\triangleright$ Equal pairs of sides are the two lateral sides of an isosceles triangle, as well as the two sides adjacent to each other along the cut line. Equal angles are the angles at the base of the isosceles triangle. $\triangleleft$
This problem reminds us that in the first criterion for the congruence of tria... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,912 |
100. The center of a circle and any two points on it are vertices of an isosceles triangle. | $\Delta$ This is a very simple problem: two sides of the resulting triangle are radii and therefore equal. $\triangleleft$ | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,913 |
101. Point $X$ lies inside three circles of radius 1. Prove that their centers can be covered by a circle of radius 1. | $\Delta$ Let's draw a circle of radius 1 centered at point $X$ (depicted with a dashed line in the figure). By the condition, point $X$ lies inside the circles. Therefore, the distance from $X$ to the centers of the circles is less than 1. Hence, these centers will fall inside the drawn circle. $\triangleleft$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,914 |
102. Two circles have radii of 3 and 5, the distance between their centers is 9. Will they intersect? The same question if the distance between their centers is 1. | $\Delta$ There will be no intersection. In the first case: if $X$ were a common point of the two circles, then the distances from $X$ to the centers would be 3 and 5, and by the triangle inequality, the distance between the centers would be no more than 8.
In the second case, the smaller circle lies entirely inside th... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,915 |
103. Two villages are 3 kilometers apart. Draw where a picnic can be set up if it is required that (a) the distance to the nearest village is no more than 2 kilometers; (b) the distance to any village is no more than 2 kilometers. | $\triangleright$ For each village, there is its own circle - in it are the points that are no more than 2 kilometers away from it. In part (a), we can use points from any circle, and the answer is (as mathematicians say) the union of the circles. In part (b), we need only the points that fall into both circles, and the... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,916 |
104. The length of a chord does not exceed twice the radius of the circle (and equals it only if the chord is a diameter). | $\Delta$ Connect the ends of the chord with the center of the circle. We get a triangle formed by two radii and the chord. By the triangle inequality, the chord does not exceed the sum of the other two sides, that is, twice the radius. Equality is possible if the triangle degenerates into a segment, that is, if the cen... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,917 |
105. Prove that the segment connecting the midpoint of a chord with the center of the circle is perpendicular to the chord. | $\Delta$ This segment is the median of an isosceles triangle formed by two radii and a chord, and therefore it is its height. $\triangleleft$
## A Few More Problems | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,918 |
113. From a given point $A$ on a given line $l$, lay off a segment equal to a given one.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
113. From a given point $A$ on a given line $l$, lay off a segment equal to a given one. | $\triangle$ For this, a compass is sufficient: set the compass to the length of the given segment and construct a circle of this radius centered at A. It will intersect the line $l$ at two points, let's call them $B$ and $C$. The segments $A B$ and $A C$ will be the desired ones. $\triangleleft$
If the ruler had divis... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,919 |
114. Given a triangle $A B C$ and a segment $A^{\prime} B^{\prime}$, equal to the segment $A B$. Construct a triangle $A^{\prime} B^{\prime} C^{\prime}$, equal to the triangle $A B C$. | $\triangle$ The positions of points $A^{\prime}$ and $B^{\prime}$ are given to us. We need to find point $C^{\prime}$. The distance from it to point $A^{\prime}$ is known: it is equal to $A C$. Therefore, point $C^{\prime}$ lies on the circle with center at $A^{\prime}$ and radius $A C$. Let's construct this circle. Si... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,920 |
116. From the given ray, lay off an angle equal to the given one.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | $\checkmark$ This problem can be reduced to the previous ones. Take any two points on the sides of the angle, and
you will get a triangle. Construct an equal triangle (by drawing two circles... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,922 |
117. Find the midpoint of the given segment. | $\checkmark$ We will construct the perpendicular bisector of the segment, thereby finding its midpoint. We know that the points on the perpendicular bisector are equidistant from the endpoints of the segment. Therefore, it is sufficient to find two such points and draw a line through them. To find them, we will constru... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,923 |
118. Bisect the given angle. | $\triangleright$ Let $O$ be the vertex of the angle. Using a compass, we can lay off equal segments $OA$ and $OB$ on its sides. The length of these segments (i.e., the compass setting) can be any. We already know from the solution to the previous problem how to construct the perpendicular bisector of segment $AB$. Poin... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,924 |
119. Given a point $A$ on a line $l$. Draw a line through $A$ perpendicular to $l$. | $\triangleright$ This problem can be easily reduced to already solved ones: we know how to construct the perpendicular bisector of a segment (solution to problem 117), so it is sufficient to find a segment on line $l$ whose midpoint is point $A$. This can be easily done by laying off equal segments from $A$ in opposite... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,925 |
120. Given a point $A$, not lying on a line $l$. Draw a line through $A$ that is perpendicular to $l$. | $\triangleright$ This problem can also be reduced to the previous ones: if we draw a circle centered at $A$, intersecting the line at two points, then these points of intersection are equidistant from $A$, and therefore $A$ lies on the perpendicular bisector of the segment with endpoints at these points. And we know ho... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,926 |
124. Prove that two lines parallel to a third line are parallel to each other. | $\triangle$ This problem can be solved in two ways, either using statement (1) or using
statement (2). We will present both options.
, since the angles formed by the intersection of two lines with their common perpendicular are right and therefore equal. $\triangleleft$
Pairs of angles formed by the intersection of two lines by a third (traditionally called a transversal) have traditional names, ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,928 |
126. Prove that through a given point, only one line can be drawn perpendicular to a given line. | $\Delta$ This is an exact repetition of the previous problem: if two such lines are drawn, they will be two perpendiculars to the given line. From the previous problem, they are parallel - and cannot pass through one point (without coinciding)! $\triangleleft$
There are two expressions concerning the construction of p... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,929 |
127. In quadrilateral $A B C D$, angle $A C B$ is equal to angle $C A D$. Prove that it has two parallel sides. | $\triangle$ This problem is a direct consequence of property (2): by the condition, the alternate interior angles $ACB$ and $CAD$ formed by the transversal $AC$ are equal, and therefore the lines $AD$ and $BC$ are parallel. $\triangleleft$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,930 |
128. Two adjacent angles of a quadrilateral sum up to $180^{\circ}$. Prove that two of its sides are parallel. Prove that the other two angles also sum up to $180^{\circ}$. | $\triangleright$ Let in quadrilateral $A B C D$ angles $A$ and $B$ sum up to $180^{\circ}$. They are consecutive interior angles formed by the transversal $A B$, and therefore lines $A D$ and $B C$ are parallel.
Now consider side $C D$ as the transversal, where angles $C$ and $D$ are consecutive interior angles. Since... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,931 |
130. The sides of one angle are parallel to the sides of another (and run in the same directions, see figure). Prove that the angles are equal. | $\triangleright$ Both these angles are equal to the third one, which is formed by the intersection of one side of one angle with the non-parallel side of the other angle (we use property (2) twice). $\triangleleft$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,933 |
131. Through the vertex $B$ of triangle $A B C$, a line parallel to side $A C$ is drawn. Mark on the diagram the angles equal to the angles $A$ and $C$ of the triangle. From this, deduce that the sum of the angles of the triangle is $180^{\circ}$. | $\triangleright$ The angles marked in the figure are equal to angles $A$ and $C$ as alternate interior angles and supplement angle $B$ from both sides to form a straight angle. $\triangleleft$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,934 |
132. In a triangle, all sides are equal. Find its angles. | $\Delta$ An equilateral triangle (in which all sides are equal) is a special case of an isosceles triangle, and any side can be taken as the base. Therefore, all angles in it are equal. Since the angles sum up to $180^{\circ}$ (from the previous problem), each of them is $60^{\circ} . \triangleleft$ | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,935 |
133. In an isosceles triangle, one of the angles is $80^{\circ}$. What are its other angles? (List all possibilities.) | $\triangleright$ In an isosceles triangle, the two angles at the base are equal. If one of them is $80^{\circ}$, then together they sum up to $160^{\circ}$, and the third angle (at the vertex) is left with $20^{\circ}$. If, however, the angle at the vertex is $80^{\circ}$, then the two angles at the base are left with ... | 20,50 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,936 |
134. In an isosceles triangle, one of the angles is $60^{\circ}$. Prove that the triangle is equilateral. | $\triangle$ It doesn't matter which angle is $60^{\circ}$ (whether it's at the base or at the vertex), the other angles will still be $60^{\circ}$. But we know that if two angles in a triangle are equal, then the sides opposite them are also equal. Therefore, since all three angles are equal, all three sides are also e... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,937 |
135. What values can the largest angle of a triangle take? What values can the smallest angle of a triangle take? What values can the third (middle in size) angle of a triangle take? | $\triangleright$ If three numbers sum up to 180, then the largest of them is not less than 60 (otherwise, all the others would be less than 60, and the sum would be less than 180). Thus, the largest angle in a triangle can be from $60^{\circ}$ to $180^{\circ}$. By similar reasoning, the smallest angle can be from $0^{\... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,938 |
136. Prove that the sum of the angles of a quadrilateral is $360^{\circ}$ by cutting it into two triangles. | $\triangle$ By drawing a diagonal, we cut a quadrilateral into two triangles. In this process, two angles of the quadrilateral are divided into two parts (each). If we sum all six angles in the two triangles, we get $2 \times 180^{\circ}=360^{\circ}$. On the other hand, this sum represents the sum of the angles of the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,939 |
141. What is the sum of the exterior angles at three vertices of a triangle? | $\Delta$ Together with the three interior angles (which sum up to $180^{\circ}$), this results in three straight angles, that is, $540^{\circ}$, so the sum of the exterior angles is $540-180=360^{\circ} . \triangleleft$ | 360 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,940 |
142. What is the sum of the exterior angles at the four vertices of a convex quadrilateral? | $\Delta$ To find the sum of these angles (they are shown on the diagram with single arcs, but this does not mean they are equal!), consider the corresponding interior angles. Together with the 4 interior angles (which sum up to $360^{\circ}$), we get 4 straight angles, that is, $720^{\circ}$, so the sum of the exterior... | 360 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,941 |
143. What is the sum of the exterior angles at the five vertices of a convex pentagon? | $\checkmark$ Together with the 5 interior angles (which sum up to $540^{\circ}$), we get 5 straight angles, that is, $900^{\circ}$, so the sum of the exterior angles is $900-540=360^{\circ} . \triangleleft$
Similarly, one can verify that the sum of the exterior angles of a convex hexagon, heptagon, and generally any $... | 360 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,942 |
144. Given a straight line and a point not lying on it. How to draw (with a compass and a ruler) a straight line through the given point, parallel to the given line? | $\triangleright$ Here is one way. Let a line $l$ and a point $A$ be given. Draw any line $m$ through point $A$ that intersects line $l$. The desired line intersects line $m$ at the same angle as line $l$, so it is simply necessary to lay off this angle from line $m$ (Problem 116).
We can choose the angle of intersecti... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,943 |
160. Prove that in a right triangle, the angles at the hypotenuse sum up to a right angle. | $\triangle$ The sum of all three angles is $180^{\circ}$, so the remaining for the other two is $180-90=90^{\circ} . \triangleleft$
From this, it follows that the other angles in a right triangle (except the right angle) are acute: if their sum equals a right angle, then each of the addends is less than a right angle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,944 |
161. Find the angles of a right isosceles triangle. | $>$ The angles at the base of an isosceles triangle are equal, and together they sum up to $90^{\circ}$, so each of them is $45^{\circ} . \triangleleft$
Several of the following problems provide criteria for the congruence of right triangles. | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,945 |
162. (a) Prove the equality criterion for right triangles "by two legs": if in two right triangles ($ABC$ and $A'B'C'$ with right angles at $B$ and $B'$) the corresponding legs are equal ($AB = A'B'$ and $BC = B'C'$), then the triangles are equal.
(b) Formulate and prove the equality criterion for right triangles by a... | $\triangle$ These properties do not require special proof: they are merely special cases of the first and second criteria for the congruence of triangles, when one of the angles is a right angle. $\triangleleft$
In the next problem, it is no longer possible to refer to the criteria for the congruence of triangles (bec... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,946 |
163. The leg and hypotenuse of one right triangle are equal to the leg and hypotenuse of another. Prove that the triangles are congruent.
 | $\triangle$ Place the triangles against each other with the equal legs $B C$ and $B^{\prime} C^{\prime}$ (see the figure). Since the triangles are right-angled, the other two legs will form a single straight line $\left(A A^{\prime}\right)$, and the two triangles will form a larger triangle. By the condition, the hypot... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,947 |
164. The hypotenuse of one right-angled triangle is equal to the hypotenuse of another; one of the angles adjacent to it is equal to the angle at the hypotenuse of the other triangle. Prove that the triangles are equal. | $\triangleright$ Since the sum of the angles in both triangles is $180^{\circ}$ and they have two equal angles, the third angles are also equal, and the triangles are congruent by the hypotenuse and the two adjacent angles. | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,948 |
165. The leg of one right triangle is equal to the leg of another; the angles opposite them are also equal. Prove that the triangles are equal. | $\triangleright$ In each of the two triangles, the acute angles sum up to $90^{\circ}$, so if the angles opposite the equal legs are equal, then the angles adjacent to them are also equal. It remains to use the second criterion for the congruence of triangles. $\triangleleft$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,949 |
167. In a right-angled triangle, one of the angles is $30^{\circ}$. Prove that one of the legs is half the length of the hypotenuse. | $\triangleright$ Let's apply two such triangles to each other by their legs so that the two $30^{\circ}$ angles form a $60^{\circ}$ angle. This will result in an isosceles triangle with a vertex angle of $60^{\circ}$, which will be equilateral. Therefore, the doubled shorter leg
, and angles $B$ and $D$ are supplementary to a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,953 |
174. Prove that the diagonal divides a parallelogram into two equal triangles. | $\triangle$ In parallelogram $A B C D$, draw diagonal $B D$. Since sides $A B$ and $C D$ are parallel, angles $A B D$ and $B D C$ are equal as alternate interior angles. Similarly, the parallelism of sides $B C$ and $A D$ ensures that angles $C B D$ and $B D A$ are equal. Diagonal $B D$ is a common side of triangles $A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,954 |
176. The diagonals divide the parallelogram into 4 triangles. Prove that there are two pairs of equal ones. | $\triangleright$ Let $O$ be the point of intersection of the diagonals of parallelogram $A B C D$. We will prove that triangles $A O D$ and $C O B$ are equal. Indeed, sides $A D$ and $B C$ are equal (by the previous problem), angles $O A D$ and $O C B$, as well as $O D A$ and $O B C$ are equal as alternate interior ang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,956 |
178. Prove that if in a quadrilateral $A B C D$ the opposite sides are pairwise equal ($A B=C D, A D=B C$), then it is a parallelogram. | $\checkmark$ Let's draw the diagonal $B D$. It will divide the quadrilateral into two triangles that are equal by the three sides ($B D$ is common, the other pairs of sides are equal by assumption). Therefore, angles $C D B$ and $A B D$ are equal, so lines $A B$ and $C D$ are parallel (the angles are alternate interior... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,958 |
179. Prove that a quadrilateral in which opposite angles are equal in pairs is a parallelogram | $\checkmark$ Let two opposite angles of a quadrilateral be equal to $\alpha$, and the other two angles be equal to $\beta$. Then the sum of all angles, that is, $2 \alpha+2 \beta$, is $360^{\circ}$, so $\alpha+\beta=180^{\circ}$. Therefore, the opposite sides of the quadrilateral are parallel (angles $\alpha$ and $\bet... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,959 |
180. The point of intersection of the diagonals of a quadrilateral divides each of them in half. Prove that this quadrilateral is a parallelogram. | $\triangleright$ Let $O$ be the point of intersection of the diagonals of quadrilateral $ABCD$. Triangles $AOB$ and $COD$ are congruent by the first criterion (angles at vertex $O$ are vertical, sides are equal by condition). In particular, angles $OAB$ and $OCD$ are equal. These angles are alternate interior angles wi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,960 |
181. Prove that if in quadrilateral $A B C D$ sides $A D$ and $B C$ are equal and parallel, then it is a parallelogram | $\triangleright$ Let's draw the diagonals; let $O$ be their point of intersection. Consider triangles $A O D$ and $B O C$. They are congruent by a side and two angles ($A D=B C$, the adjacent angles are equal as alternate interior angles), so $A O=O C, B O=O D$. Now we can refer to the previous problem (or repeat the r... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,961 |
182. Two parallelograms $A B C D$ and $D C E F$ are attached to each other along the common side $C D$. Prove that the quadrilateral $A B E F$ is also a parallelogram. | $\triangleright$ Let's look at the segments $A B, D C$ and $F E$. Since $A B C D$ is a parallelogram, the first segment is parallel and equal to the second. Since $D C E F$ is a parallelogram, the second segment is equal and parallel to the third. Therefore, the first and third segments are equal and parallel (recall t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,962 |
183. In triangle $ABC$, the median $AM$ is extended beyond point $M$ to point $D$ at a distance equal to $AM$ (so that $AM = MD$). Prove that $ABDC$ is a parallelogram. | $\triangleright$ In the resulting quadrilateral, the diagonals are bisected by the point of intersection by construction, it remains to use problem 180. $\triangleleft$
Such a construction (extending the median by twice) has already been encountered by us. With its help, we proved that the median in a triangle does no... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,963 |
184. A line passes through the center of a parallelogram. Prove that the segment of the line enclosed within the parallelogram is bisected by the center. | $\triangleright$ In the shaded triangles, the sides that are halves of the diagonal are equal (the center divides the diagonal in half), and the angles adjacent to them are equal (one pair are vertical angles, the other pair are alternate interior angles). $\triangleleft$
## A few more problems | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,964 |
195. If a parallelogram has at least one right angle, then it is a rectangle. | $\triangle$ In a parallelogram, opposite angles are equal, and the sum of adjacent angles is $180^{\circ}$. Therefore, if one of the angles is $90^{\circ}$, then the adjacent and opposite angles will also be $90^{\circ} . \triangleleft$ | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,965 |
196. A quadrilateral in which all angles are right angles is a rectangle. | $\triangleright$ We need to prove that this is a parallelogram. Indeed, the opposite sides are parallel, since the sum of the interior angles on the same side is equal to two right angles. $\triangleleft$ | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,966 |
197. Prove that the diagonals of a rectangle are equal.
 | $\triangle$ Let's cut this parallelogram into two triangles in two ways (by drawing two different diagonals). All four triangles will be right-angled and will have the same legs (the opposite sides of the rectangle, like any parallelogram, are equal). Therefore, the triangles are equal, and their hypotenuses are also e... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,967 |
198. Prove that if the diagonals of a parallelogram are equal, then it is a rectangle. | $\triangleright$ As in the previous problem, let's cut it into triangles in two ways. Again, all these triangles are equal, but for a different reason: by three sides (pairs of opposite sides of the parallelogram and its diagonal). Therefore, all angles of the parallelogram are equal. Together they sum up to $360^{\cir... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,968 |
199. Prove that in a right-angled triangle, the median drawn from the vertex of the right angle is equal to half the hypotenuse.
64 | $\triangleright$ Again, we will apply the technique of doubling the median. Let $ABC$ be a right triangle, with angle $A$ being the right angle, and $AM$ being the median (so $M$ is the midpoint of side $BC$). Extend $AM$ beyond point $M$ by a distance equal to $AM$. We obtain point $N$, which we connect to vertices $B... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,969 |
200. A ladder is leaning against a vertical wall and starts to slide down. Along what line does a cat sitting at the midpoint of the ladder move? | $\Delta$ Along the arc of a circle. Indeed, the ladder is the hypotenuse of a right triangle, and the cat is the midpoint of the hypotenuse, so the distance from the cat to the right-angle vertex is half the length of the ladder and does not change. Therefore, the cat moves along an arc (part) of a circle. $\trianglele... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,970 |
201. Perpendiculars are dropped from two points on a straight line to a line parallel to it. Prove that they are equal. | $\Delta$ Indeed, it turns out to be a quadrilateral with all right angles, that is, a rectangle. Its opposite sides are equal. $\triangleleft$
Thus, the distance between two parallel lines can be determined as the length of the segment cut by them on a perpendicular line - according to the previous problem, it does no... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,971 |
202. Perpendiculars dropped from points $A$ and $B$ to line $l$ are equal. Prove that segment $A B$ is parallel to line $l$.
 | $\triangleright$ In the resulting quadrilateral, two sides (the perpendiculars dropped onto $l$) are equal and parallel, so it is a parallelogram, and the other two sides are also parallel. $\triangleleft$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,972 |
205. The diagonals of a rhombus intersect at right angles. | $\Delta$ Let's cut it into four triangles with its diagonals. They will turn out to be equal by three sides (let's recall that the diagonals of any parallelogram are bisected by the point of intersection). Therefore, all four angles formed at the intersection of the diagonals are equal. Therefore, they are right angles... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,975 |
206. If in a parallelogram the diagonals intersect at a right angle, then it is a rhombus. | $\triangleright$ As in the previous problem, let's cut the parallelogram into four triangles using its diagonals. They are equal again, but this time not by three sides, but by two sides and the (right) angle between them. $\triangleleft$
|| A rectangle that is also a rhombus is called a square: all its sides are equa... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,976 |
207. What angle does the diagonal of a square form with its side? | $\Delta$ This is an acute angle in a right isosceles triangle, which equals $45^{\circ} . \triangleleft$
## A few more problems | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,977 |
240. Three vertices of a parallelogram (in order of traversal) have coordinates $(a, b),(0,0)$ and $(c, d)$. What are the coordinates of the fourth vertex? | $\Delta$ We have already seen that if we count the same number of cells horizontally and vertically, we get an equal and parallel segment. Therefore, we need to count from the point $(c+d)$ to the right by $a$ cells and up by $b$ cells. This will result in the point $(c+a, d+b) . \triangleleft$
Physicists call this op... | (+,+b) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,981 |
241. A line is drawn through the points $(0,0)$ and $(5,3)$. Does it pass through any other grid nodes? | $\triangleright$ Of course, it passes. From the node $(5,3)$, let's lay out five more cells to the right and three cells up. This
segment will have the same slope as the segment $(0,0)-(5,3... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,982 |
242. Can a straight line be drawn on graph paper that does not pass through any grid node? | $\checkmark$ It is possible: we draw a line with an inclination angle of $45^{\circ}$ that passes through the grid nodes, and then shift it half a cell upwards (see the figure). It is easy to notice that it intersects all the horizontals and verticals at the midpoints of the cells. $\triangleleft$
The more interesting... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,983 |
263. Using a compass, draw a circle. Then, without changing the compass setting, lay out the distance equal to its radius six times on the circle (the distance is measured not along the circumference, but along the chord). If done carefully, the last point will coincide with the starting point. Why? | $\Delta$ Two adjacent points and the center of the circle are the vertices of a triangle (its sides are radii and a chord), and this triangle is equilateral. Therefore, all angles are $60^{\circ}$, and six such angles will form a full circle. $\triangleleft$ | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,984 |
265. Inside the square $A B C D$, an equilateral triangle $A B M$ is constructed, and vertex $M$ is connected to points $C$ and $D$. Find the angles of triangle $C M D$. | $\Delta$ The angles of an equilateral triangle are $60^{\circ}$, so the angle at vertex $A$ of the isosceles triangle $A M D$ is $90-60=30^{\circ}$. Therefore, the angles at its base are $(180-30) / 2=75^{\circ}$. Thus, the angle $C D M$ is $90-75=15^{\circ}$ (as well as $\angle M C D$ ). $\triangleleft$ | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,986 |
266. Inside the square $A B C D$, an isosceles triangle $A B L$ is constructed with base $A B$ and base angles of $15^{\circ}$. Under what angle is the side $C D$ seen from the vertex $L$? | $\triangleright$ Of course, this problem essentially coincides with the previous one: construct an equilateral triangle $C D M$ inside the square, then $A M$ and $B M$ will exactly form a $15^{\circ}$ angle with the sides, so $M$ will coincide with $L . \triangleleft$
If you don't know the previous problem, this one
... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,987 |
267. In a hexagon, all angles of which are equal to $120^{\circ}$, the sides are equal to $a, b, c, d, e, f$ (clockwise). Prove that
$$
d-a=b-e=f-c
$$ | $\checkmark$ Let's first prove that the opposite sides are parallel. Extend sides $a$ and $c$ until they intersect. In the resulting triangle, all angles are $60^{\circ}$. Now it is clear that sides $c$ and $f$ are parallel because the interior alternate angles formed by the transversal $a$ are $120^{\circ}$ and $60^{\... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,988 |
278. Through the midpoint $M$ of side $B C$ of triangle $A B C$, lines are drawn parallel to sides $A B$ and $A C$. Prove that they intersect sides $A C$ and $A B$ at their midpoints. | $\triangleright$ These lines cut the triangle into two triangles $B M K$ and $M C L$ and a quadrilateral $A K M L$, which by construction is a parallelogram. Triangles $B M K$ and $M C L$ are equal by sides $B M=M C$ and the angles adjacent to them (which are equal as angles formed by a transversal). In particular, $K ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,989 |
279. A segment drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side. | $\triangleright$ This is exactly what was claimed in the previous problem (for two such segments). $\triangleleft$
(We deliberately asked about two segments first, as it is more convenient to prove this for two segments simultaneously.) | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,990 |
280. Prove that the segment connecting the midpoints of two sides ($AB$ and $BC$) of triangle $ABC$ is parallel to the third side ($AC$) and half its length. | $\triangle$ In the previous problem, we drew a segment through the midpoint $M$ of side $BC$ parallel to side $AC$ and verified that it bisects side $AB$. Now, we have bisected side $AB$ and want to confirm that the segment connecting the midpoints of the sides is parallel to side $AC$. But this is the same thing: both... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,991 |
281. Any triangle can be cut into 4 equal triangles. How? | $\Delta$ Draw three midlines - in all the resulting triangles, the sides will be equal to half the sides of the original, and therefore all four tri-
angles will be equal. $\triangleleft$ | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,992 | |
283. Prove that the midpoints of the sides of a rhombus form a rectangle, the midpoints of the sides of a rectangle form a rhombus, and the midpoints of the sides of a square form a square. | $\Delta$ (We have already encountered these statements earlier, in problems 215 and 216, but now we can reason more simply.) The solution to the previous problem shows that the sides of the parallelogram are parallel to the diagonals of the quadrilateral and equal to their halves. Therefore, if the diagonals are perpen... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,994 |
285. Given three arbitrary points on a plane, not lying on the same line. Construct a triangle for which they will be the midpoints of the sides. | $\triangleright$ The segments connecting the given points ($A, B, C$) will be the midlines of the desired triangle, so its sides can be easily constructed (by drawing a line through $A$ parallel to $BC$ and so on). This will result in 4 equal triangles (they are equal because the diagonal of a parallelogram divides it ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,995 |
286. In a triangle, two medians are drawn. Prove that the point of their intersection divides each median in the ratio $2: 1$ (counting from the vertex). | $\triangleright$ Let in triangle $ABC$ the medians $AK$ and $CL$ intersect at point $O$. We need to prove that $AO: OK = CO: OL = 2: 1$. Mark the midpoints of segments $AO$ (let this be point $M$) and $CO$ (point $N$). We need to prove that each median is divided into three equal parts.
Draw segments $LK$ and $MN$. Ea... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,996 |
287. Prove that the three medians of a triangle intersect at one point. | $\triangleright$ In essence, we have already proven this in the previous problem. Indeed, let's draw one of the medians. At what point will the second median intersect it? We know this: at the point that divides it in the ratio $2: 1$ (at a distance of one-third from the end). And where will the third median intersect ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,997 |
## 16. Thales' Theorem
Let there be three lines $l, m$, and $n$. Consider a point on line $l$ that casts a shadow on line $m$ (the "screen") when light falls parallel to line $n$. In other... | $\checkmark$ Draw segments $A B_{2}$ and $C D_{2}$ parallel to line $m$. They are parallel to each other, and therefore in triangles $A B B_{2}$ and $C D D_{2}$, angles $A$ and $C$ are equal. Lines $B B_{1}$ and $D D_{1}$ are also parallel, so angles $B$ and $D$ in these triangles are equal. Consequently, the triangles... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,998 |
294. Points $A, B, C$ lie on line $l$, points $A_{1}, B_{1}, C_{1}$ are their shadows on $m$. Point $B$ divides segment $A C$ in the ratio $2: 3$. Prove that point $B_{1}$ divides segment $A_{1} C_{1}$ also in the ratio $2: 3$. | $\triangleright$ It is sufficient to divide the segment $A C$ into five equal parts, two of which will constitute $A B$, and three will constitute $B C$, and draw through the points of division lines parallel to the line $n$. According to Thales' theorem, they will cut five equal segments on the line $m$, two of which ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,000 |
295. If two segments on a line $l$ are in the ratio $5: 7$, then their shadows on a line $m$ are also in the ratio $5: 7$. | $\triangleright$ The difference in this problem from the previous one is that the segments on the line $l$ do not necessarily touch each other; they can be in different places (or even intersect). However, the solution remains essentially the same. We will divide the first segment into 5 parts, and the second into 7 pa... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,001 |
296. Derive from Thales' theorem the known property of the midline of a triangle: if in triangle $ABC$ from the midpoint $D$ of side $AB$ a segment is drawn parallel to $AC$ until it intersects $BC$ at point $E$, then this point will be the midpoint of $BC$. | $\triangleright$ Points $B, E$ and $C$ are the shadows of points $B, D$ and $A$ when projected (illuminated) along the line $A C$. It remains to apply Thales' theorem to the equal segments $A D$ and $D B$ and conclude that their shadows $C E$ and $E B$ are equal. (The fact that point $B$ is its own shadow does not inte... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,002 |
298. On the sides $A B$ and $C B$ of triangle $A B C$, points $D$ and $E$ are taken, dividing them in the ratio $A D: D B = C E: E B = 2: 1$. Prove that $D E$ is parallel to $A C$. | $\triangleright$ If $DE$ is not parallel to $AC$, draw another line through $D$, parallel to $AC$, until it intersects side $BC$ at point $E'$. By the previous problem, point $E'$ divides side $CB$ in the same ratio $2:1$ as point $E$, and therefore coincides with $E$. $\triangleleft$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,004 |
300. A segment on the line $l$ is three times longer than its shadow. Prove that any other segment on the line $l$ is also three times longer than its shadow. | $\Delta$ This follows from the general theorem of Thales and the properties of proportions. Indeed, let the segment be three times longer than its shadow. If the length of the segment is increased by some factor, then by the theorem of Thales, its shadow will also increase by the same factor, and the segment will remai... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,006 |
315. Two of the angles of a trapezoid are $60^{\circ}$ and $130^{\circ}$. What can be said about its other angles? | $\Delta$ The angles at the lateral side of the trapezoid add up to $180^{\circ}$ (as interior alternate angles). Therefore, there are also angles $180^{\circ}-60^{\circ}=120^{\circ}$ and $180^{\circ}-130^{\circ}=50^{\circ}$ in the trapezoid. Thus, all four angles have been found. $\triangleleft$ | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,008 |
316. Two of the angles of a trapezoid are $60^{\circ}$ and $120^{\circ}$. What can be said about its other angles? | $\triangleright$ The difference from the previous problem is that now these angles sum up to $180^{\circ}$ and therefore can quite well be the angles at the same lateral side of a trapezoid. Then the angles at the other side of the trapezoid can be arbitrary, as long as their sum is $180^{\circ} . \triangleleft$ | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 43,009 |
317. Prove that the segment connecting the midpoints of the lateral sides of a trapezoid is parallel to its bases | $\triangle$ We will act "in the opposite direction": draw a segment through the midpoint of the lateral side, parallel to the base, and convince ourselves that it will bisect the other side (and thus coincide with the segment from the problem's condition). But this is a direct consequence of Thales' theorem. $\triangle... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,010 |
318. Prove that the midline of a trapezoid is equal to half the sum of its bases. (Hint. Draw a diagonal of the trapezoid.)
 | $>$ Let's draw the diagonal $A C$ of the trapezoid $A B C D$.
Then, through the midpoint of the lateral side $A B$, draw a line parallel to the bases of the trapezoid. It will intersect th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,011 |
319. Prove that in any quadrilateral $ABCD$, the distance between the midpoints of opposite sides $AB$ and $CD$ does not exceed the half-sum $(BC + AD) / 2$ and that this inequality becomes an equality only for trapezoids. | $\triangleright$ Let's draw the diagonal $A C$. Let $M$ be its midpoint. The point $M$ together with the midpoints of sides $A B$ and $C D$ forms a triangle, one side of which is half the length of base $B C$ (as the midline), another is half the length of base $A D$, and the third is the distance in question in the pr... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 43,012 |
320. Prove that in an isosceles trapezoid, the angles at the base are equal. | $\triangleright$ Draw a line through vertex $C$ of trapezoid $A B C D$ parallel to the lateral side $A B$. It will cut the trapezoid into parallelogram $A B C M$ and triangle $C D M$. In this case, $C M$ will be equal to $A B$, and angle $C M D$ will be equal to angle $A$ of the trapezoid. If the trapezoid is isosceles... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,013 |
322. Prove that in an isosceles trapezoid, the diagonals are equal. | $\triangle$ We already know that the angles at the base $A D$ of the isosceles trapezoid $A B C D$ are equal, so triangles $A B D$ and $A C D$ are equal by two sides and the angle between them. $\triangleleft$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,015 |
323. Prove that if the diagonals of a trapezoid are equal, then it is isosceles. | $\Delta$ This task is a bit more complicated than the previous one, here additional construction is required. Draw the line $C E$ parallel to the diagonal $B D$ until it intersects the base at point $E$. In the parallelogram $B C E D$, the opposite sides $B D$ and $C E$ are equal, so $A C = C E$. In the isosceles trian... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,016 |
328. Prove that in a triangle, the larger angle lies opposite the longer side. (In other words, if in triangle $ABC$ side $AB$ is longer than side $AC$, then angle $C$ is greater than angle $B$.) | $\triangleright$ Let's mark off a segment $A D$ on side $A B$, equal to $A C$. In the isosceles triangle $A C D$, the angles at the base are equal. It remains to note that these equal angles are less than angle $A C B$ (angle $A C D$ is part of it), but greater than angle $A B C$ (in triangle $B C D$, angle $B$ is an i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,017 |
329. Prove that in a triangle, the side opposite the larger angle is longer. (In other words, if in triangle $ABC$ angle $C$ is greater than angle $B$, then $AB$ is longer than $AC$.) | $\triangleright$ In essence, this problem is already solved. We know that the larger angle lies opposite the longer side and that equal angles lie opposite equal sides (property of an equilateral triangle). Let's compare sides $A B$ and $A C$. Can they be equal? They cannot, because then angles $B$ and $C$ would be equ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,018 |
331. Given a line $l$ and a point $A$ not lying on it. Prove that the perpendicular dropped from the point to the line is the shortest distance from point $A$ to points lying on the line $l$. | $\triangleright$ We need to compare the perpendicular $A M$ with the distance $A N$ from point $A$ to some other point $N$ on the line $l$. But we already know that in the right triangle $A M N$, the hypotenuse $A N$ is longer than the leg $A M . \triangleleft$
The statement of the previous problem is sometimes formul... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,020 |
332. The line $l$ is the perpendicular bisector of the segment $A B$. Prove that points lying on one side of this line are closer to $A$ than to $B$, and on the other side - the opposite. | $\triangle$ We have already seen that points on the very perpendicular bisector are equally distant from $A$ and from $B$. Now consider a point $X$, located on the same side as point $A$, connect it with $A$ and $B$ by segments, and show that $X A < X B$. The segment $X B$ intersects the perpendicular bisector at some ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,021 |
333. Prove that if $X$ lies on the bisector, then the perpendiculars are equal; if $X$ lies on the same side of the bisector as side $O B$, then the perpendicular dropped to $O B$ is shorter; if $X$ lies on the other side of the bisector, the other perpendicular is shorter. | $\triangleright$ Drop perpendiculars $X P$ and $X Q$ from point $X$ to the sides $O A$ and $O B$ of the given angle. If $X$ lies on the angle bisector, then the right triangles $O X P$ and $O X Q$ are congruent by the hypotenuse and acute angle.
Now suppose $X$ does not lie on the bisector and, for instance, lies on t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,022 |
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