Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$. Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.

A circle $C$ with center $O$ on base $BC$ of an isosceles triangle $ABC$ is tangent to the equal sides $AB,AC$. If point $P$ on $AB$ and point $Q$ on $AC$ are selected such that $PB \times CQ = (\frac{BC}{2})^2$, prove that line segment $PQ$ is tangent to circle $C$, and prove the converse.

Two different points $X$ and $Y$ are chosen in the plane. Find all the points $Z$ in this plane for which the triangle $XYZ$ is isosceles.

Let $ABC$ be a triangle, and $M$ an interior point such that $\angle MAB=10^\circ$, $\angle MBA=20^\circ$, $\angle MAC=40^\circ$ and $\angle MCA=30^\circ$. Prove that the triangle is isosceles.

On the sides of triangle $ABC$, isosceles right-angled triangles $AUB, CVB$, and $AWC$ are placed. These three triangles have their right angles at vertices $U, V$ , and $W$, respectively. Triangle $AUB$ lies completely inside triangle $ABC$ and triangles $CVB$ and $AWC$ lie completely outside $ABC$. See the figure. Prove that quadrilateral $UVCW$ is a parallelogram. [asy] import markers; unitsize(1.5 cm); pair A, B, C, U, V, W; A = (0,0); B = (2,0); C = (1.7,2.5); U = (B + rotate(90,A)*(B))/2; V = (B + rotate(90,C)*(B))/2; W = (C + rotate(90,A)*(C))/2; draw(A--B--C--cycle); draw(A--W, StickIntervalMarker(1,1,size=2mm)); draw(C--W, StickIntervalMarker(1,1,size=2mm)); draw(B--V, StickIntervalMarker(1,2,size=2mm)); draw(C--V, StickIntervalMarker(1,2,size=2mm)); draw(A--U, StickIntervalMarker(1,3,size=2mm)); draw(B--U, StickIntervalMarker(1,3,size=2mm)); draw(rightanglemark(A,U,B,5)); draw(rightanglemark(B,V,C,5)); draw(rightanglemark(A,W,C,5)); dot("$A$", A, S); dot("$B$", B, S); dot("$C$", C, N); dot("$U$", U, NE); dot("$V$", V, NE); dot("$W$", W, NW); [/asy]

From the vertices of a regular 27-gon, seven are chosen arbitrarily. Prove that among these seven points there are three points that form an isosceles triangle or four points that form an isosceles trapezoid.

Restore the isosceles triangle $ABC$ ($AB = AC$) if the common points $I, M, H$ of bisectors, medians and altitudes respectively are given.

Let $ABC$ be an isosceles triangle with $AB = AC$. Suppose $P,Q,R$ are points on segments $AC, AB, BC$ respectively such that $AP = QB$, $\angle PBC = 90^\circ - \angle BAC$ and $RP = RQ$. Let $O_1, O_2$ be the circumcenters of $\triangle APQ$ and $\triangle CRP$. Prove that $BR = O_1O_2$. [i]Proposed by Atul Shatavart Nadig[/i]

Let $P$ be a point inside the triangle $ABC$ such that $AB = BC$, $\angle ABC = 80^o$, $\angle PAC = 40^o$ and $\angle ACP = 30^o$. Find $\angle BPC$. (G Galperin)

In a triangle $ABC$ let $E$ be the midpoint of the side $AC$ and $F$ the midpoint of the side $BC$. Let $G$ be the foot of the perpendicular from $C$ to $ AB$. Show that $\vartriangle EFG$ is isosceles if and only if $\vartriangle ABC$ is isosceles.

Let $ABC$ be a triangle. Let $D,E$ be the points outside of the triangle so that $AD=AB,AC=AE$ and $\angle DAB =\angle EAC =90^o$. Let $F$ be at the same side of the line $BC$ as $A$ such that $FB = FC$ and $\angle BFC=90^o$. Prove that the triangle $DEF$ is a right- isosceles triangle.

Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$. Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.

In $\vartriangle ABC, AB=AC=14 \sqrt2 , D$ is the midpoint of $CA$ and $E$ is the midpoint of $BD$. Suppose $\vartriangle CDE$ is similar to $\vartriangle ABC$. Find the length of $BD$.

In the scalene triangle $ABC$,$\angle ACB=60$ and $\Omega$ is its cirumcirle.On the bisectors of the angles $BAC$ and $CBA$ points $A^\prime$,$B^\prime$ are chosen respectively such that $AB^\prime \parallel BC$ and $BA^\prime \parallel AC$.$A^\prime B^\prime$ intersects with $\Omega$ at $D,E$.Prove that triangle $CDE$ is isosceles.(A. Kuznetsov)

Let $ABCD$ be a parallelogram with an acute angle at $A$. Let $G$ be a point on the line $AB$, distinct from $B$, such that $|CG| = |CB|$. Let $H$ be a point on the line $BC$, distinct from $B$, such that $|AB| =|AH|$. Prove that triangle $DGH$ is isosceles. [asy] unitsize(1.5 cm); pair A, B, C, D, G, H; A = (0,0); B = (2,0); D = (0.5,1.5); C = B + D - A; G = reflect(A,B)*(C) + C - B; H = reflect(B,C)*(H) + A - B; draw(H--A--D--C--G); draw(interp(A,G,-0.1)--interp(A,G,1.1)); draw(interp(C,H,-0.1)--interp(C,H,1.1)); draw(D--G--H--cycle, dashed); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, E); dot("$D$", D, NW); dot("$G$", G, NE); dot("$H$", H, SE); [/asy]

On the sides of triangle $ABC$, isosceles right-angled triangles $AUB, CVB$, and $AWC$ are placed. These three triangles have their right angles at vertices $U, V$ , and $W$, respectively. Triangle $AUB$ lies completely inside triangle $ABC$ and triangles $CVB$ and $AWC$ lie completely outside $ABC$. See the figure. Prove that quadrilateral $UVCW$ is a parallelogram. [asy] import markers; unitsize(1.5 cm); pair A, B, C, U, V, W; A = (0,0); B = (2,0); C = (1.7,2.5); U = (B + rotate(90,A)*(B))/2; V = (B + rotate(90,C)*(B))/2; W = (C + rotate(90,A)*(C))/2; draw(A--B--C--cycle); draw(A--W, StickIntervalMarker(1,1,size=2mm)); draw(C--W, StickIntervalMarker(1,1,size=2mm)); draw(B--V, StickIntervalMarker(1,2,size=2mm)); draw(C--V, StickIntervalMarker(1,2,size=2mm)); draw(A--U, StickIntervalMarker(1,3,size=2mm)); draw(B--U, StickIntervalMarker(1,3,size=2mm)); draw(rightanglemark(A,U,B,5)); draw(rightanglemark(B,V,C,5)); draw(rightanglemark(A,W,C,5)); dot("$A$", A, S); dot("$B$", B, S); dot("$C$", C, N); dot("$U$", U, NE); dot("$V$", V, NE); dot("$W$", W, NW); [/asy]

Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$. Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.

Let $ABC$ be a triangle, $O$ is the center of the circle circumscribed around it, $AD$ the diameter of this circle. It is known that the lines $CO$ and $DB$ are parallel. Prove that the triangle $ABC$ is isosceles. (Andrey Mostovy)

In the Triangle ABC AB>AC, the extension of the altitude AD with D lying inside BC intersects the circum-circle of the Triangle ABC at P. The circle through P and tangent to BC at D intersects the circum-circle of Triangle ABC at Q distinct from P with PQ=DQ. Prove that AD=BD-DC

Consider an isosceles triangle $ABC$ with $AB = AC$. Let $\omega(XYZ)$ be the circumcircle of the triangle $XY Z$. The tangents to $\omega(ABC)$ through $B$ and $C$ meet at the point $D$. The point $F$ is marked on the arc $AB$ of $\omega(ABC)$ that does not contain $C$. Let $K$ be the point of intersection of lines $AF$ and $BD$ and $L$ the point of intersection of the lines $AB$ and $CF$. Let $T$ and $S$ be the centers of the circles $\omega(BLC)$ and $\omega(BLK)$, respectively. Suppose that the circles $\omega(BTS)$ and $\omega(CFK)$ are tangent to each other at the point $P$. Prove that $P$ belongs to the line $AB$.

Let $ABC$ be a triangle, and $M$ an interior point such that $\angle MAB=10^\circ$, $\angle MBA=20^\circ$, $\angle MAC=40^\circ$ and $\angle MCA=30^\circ$. Prove that the triangle is isosceles.

A triangle is cut into several (not less than two) triangles. One of them is isosceles (not equilateral), and all others are equilateral. Determine the angles of the original triangle.

Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$. Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.

A set consisting of $n$ points of a plane is called an isosceles $n$-point if any three of its points are located in vertices of an isosceles triangle. Find all natural numbers for which there exist isosceles $n$-points.

Each point on the plane is colored either red, green, or blue. Prove that there exists an isosceles triangle whose vertices all have the same color.