A point $D$ is chosen on the side $AC$ of a triangle $ABC$ with $\angle C < \angle A < 90^\circ$ in such a way that $BD=BA$. The incircle of $ABC$ is tangent to $AB$ and $AC$ at points $K$ and $L$, respectively. Let $J$ be the incenter of triangle $BCD$. Prove that the line $KL$ intersects the line segment $AJ$ at its midpoint.

Starting with two points A and B, some circles and points are constructed as shown in the figure:[list][*]the circle with centre A through B [*]the circle with centre B through A [*]the circle with centre C through A [*]the circle with centre D through B [*]the circle with centre E through A [*]the circle with centre F through A [*]the circle with centre G through A[/list] [i][size=75](I think the wording is not very rigorous, you should assume intersections from the drawing)[/size][/i] Show that $M$ is the midpoint of $AB$. [img]https://cdn.artofproblemsolving.com/attachments/d/4/2352ab21cc19549f0381e88ddde9dce4299c2e.png[/img]

(a) Does there exist an innite triangular beam such that two of its cross-sections are similar but not congruent triangles? (b) Does there exist an innite triangular beam such that two of its cross-sections are equilateral triangles of sides $1$ and $2$ respectively?

Given triangle $ABC$, squares $ABEF, BCGH, CAIJ$ are constructed externally on side $AB, BC, CA$, respectively. Let $AH \cap BJ = P_1$, $BJ \cap CF = Q_1$, $CF \cap AH = R_1$, $AG \cap CE = P_2$, $BI \cap AG = Q_2$, $CE \cap BI = R_2$. Prove that triangle $P_1 Q_1 R_1$ is congruent to triangle $P_2 Q_2 R_2$.

Let $ABC$ be a triangle with incenter $I$ and $AB = 1400$, $AC = 1800$, $BC = 2014$. The circle centered at $I$ passing through $A$ intersects line $BC$ at two points $X$ and $Y$. Compute the length $XY$. [i]Proposed by Evan Chen[/i]

The winners of first AGMC in 2019 gifts the person in charge of the organiser, which is a polyhedron formed by $60$ congruent triangles. From the photo, we can see that this polyhedron formed by $60$ quadrilateral spaces. (Note: You can find the photo in 3.4 of [url]https://files.alicdn.com/tpsservice/18c5c7b31a7074edc58abb48175ae4c3.pdf?spm=a1zmmc.index.0.0.51c0719dNAbw3C&file=18c5c7b31a7074edc58abb48175ae4c3.pdf[/url]) A quadrilateral space is the plane figures that we fold the figures following the diagonal on a $n$ sides on a plane (i.e. form an appropriate dihedral angle in where the chosen diagonal is). "Two figure spaces are congruent" means they can coincide completely by isometric transform in $\mathbb{R}^3$. A polyhedron is the bounded space region, whose boundary is formed by the common edge of finite polygon. (a) We know that $2021=43\times 47$. Does there exist a polyhedron, whose surface can be formed by $43$ congruent $47$-gon? (b) Prove your answer in (a) with logical explanation.

Let $\ell$ be a tangent to the incircle of triangle $ABC$. Let $\ell_a,\ell_b$ and $\ell_c$ be the respective images of $\ell$ under reflection across the exterior bisector of $\angle A,\angle B$ and $\angle C$. Prove that the triangle formed by these lines is congruent to $ABC$.

Find all natural numbers $k$ for which there exists a set $M$ of ten real numbers such that there are exactly $k$ pairwise non-congruent triangles whose side lengths are three (not necessarily distinct) elements of $M$.

Prove that there are infinite pairs of non-congruent triangles that have the same angles and two of their equal sides. Develop an algorithm or rule to obtain these pairs of triangles and indicate at least one pair that satisfies the asserted.

Two congruent triangles are inscribed in an ellipse. Are they necessarily symmetric with respect to an axis or the center of the ellipse?

Alex dissects a paper triangle into two triangles. Each minute after this he dissects one of obtained triangles into two triangles. After some time (at least one hour) it appeared that all obtained triangles were congruent. Find all initial triangles for which this is possible.

Inside the convex pentagon $ABCDE$, a point $O$ was chosen, and it turned out that all five triangles $AOB$, $BOC$, $COD$, $DOE$ and $EOA$ are congrunet to each other. Prove that these triangles are isosceles or right-angled.

Find a right triangle that can be cut into $365$ equal triangles.

A triangle $ABC$ is given together with an arbitrary circle $\omega$. Let $\alpha$ be the reflection of $\omega$ with respect to $A$, $\beta$ the reflection of $\omega$ with respect to $B$, and $\gamma$ the reflection of $\omega$ with respect to $C$. It is known that the circles $\alpha, \beta, \gamma$ don't intersect each other. Let $P$ be the meeting point of the two internal common tangents to $\beta, \gamma$ (see picture). Similarly, $Q$ is the meeting point of the internal common tangents of $\alpha, \gamma$, and $R$ is the meeting point of the internal common tangents of $\alpha, \beta$. Prove that the triangles $PQR, ABC$ are congruent.

Two non-congruent triangles are called [i]analogous [/i] if they can be denoted as $ABC$ and $A'B'C'$ such that $AB = A'B', AC = A'C'$ and $\angle B = \angle B'$ . Do there exist three mutually [i]analogous[/i] triangles?

(B.Frenkin) Does a convex quadrilateral without parallel sidelines exist such that it can be divided into four congruent triangles?

What is the ratio of the area of the shaded square to the area of the large square? (The figure is drawn to scale) [asy] draw((0,0)--(0,4)--(4,4)--(4,0)--cycle); draw((0,0)--(4,4)); draw((0,4)--(3,1)--(3,3)); draw((1,1)--(2,0)--(4,2)); fill((1,1)--(2,0)--(3,1)--(2,2)--cycle,black);[/asy] $ \text{(A)}\ \frac{1}{6}\qquad\text{(B)}\ \frac{1}{7}\qquad\text{(C)}\ \frac{1}{8}\qquad\text{(D)}\ \frac{1}{12}\qquad\text{(E)}\ \frac{1}{16} $

Let \( ABC \) be a scalene triangle. Let \( E \) and \( F \) be the midpoints of sides \( AC \) and \( AB \), respectively, and let \( D \) be any point on segment \( BC \). The circumcircles of triangles \( BDF \) and \( CDE \) intersect line \( EF \) at points \( K \neq F \), and \( L \neq E \), respectively, and intersect at points \( X \neq D \). The point \( Y \) is on line \( DX \) such that \( AY \) is parallel to \( BC \). Prove that points \( K \), \( L \), \( X \), and \( Y \) lie on the same circle.

A point $D$ is chosen on the side $AC$ of a triangle $ABC$ with $\angle C < \angle A < 90^\circ$ in such a way that $BD=BA$. The incircle of $ABC$ is tangent to $AB$ and $AC$ at points $K$ and $L$, respectively. Let $J$ be the incenter of triangle $BCD$. Prove that the line $KL$ intersects the line segment $AJ$ at its midpoint.

Inside the convex pentagon $ABCDE$, a point $O$ was chosen, and it turned out that all five triangles $AOB$, $BOC$, $COD$, $DOE$ and $EOA$ are congrunet to each other. Prove that these triangles are isosceles or right-angled.

Given a quadrilateral, the midpoints $A, B, C, D$ of its consecutive sides, and the midpoints of its diagonals, $P$ and $Q$. Prove that $\vartriangle BCP = \vartriangle ADQ$.

Four of the eight vertices of a cube are the vertices of a regular tetrahedron. Find the ratio of the surface area of the cube to the surface area of the tetrahedron. $\text{(A)} \ \sqrt 2 \qquad \text{(B)} \ \sqrt 3 \qquad \text{(C)} \ \sqrt{\frac{3}{2}} \qquad \text{(D)} \ \frac{2}{\sqrt{3}} \qquad \text{(E)} \ 2$

In a triangle $ABC$, $D$ is a point on $BC$ such that $AD$ is the internal bisector of $\angle A$. Suppose $\angle B = 2 \angle C$ and $CD =AB$. prove that $\angle A = 72^{\circ}$.

An equilateral triangle with side length 9 is divided into 81 congruent triangles with segments which are parallel to the sides of the triangle. Prove that it cannot be divided into more than 18 parallelograms with sides 1 and 2. [I]Proposed by O.Vanyushina[/i]

A point $D$ is chosen on the side $AC$ of a triangle $ABC$ with $\angle C < \angle A < 90^\circ$ in such a way that $BD=BA$. The incircle of $ABC$ is tangent to $AB$ and $AC$ at points $K$ and $L$, respectively. Let $J$ be the incenter of triangle $BCD$. Prove that the line $KL$ intersects the line segment $AJ$ at its midpoint.