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)

On the side $AB$ of an isosceles triangle $AB$ ($AC=BC$) lie points $P$ and $Q$ such that $\angle PCQ \le \frac{1}{2} \angle ACB$. Prove that $PQ \le \frac{1}{2} AB$.

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.

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 with $\angle B > \angle C$. Let $P$ and $Q$ be two different points on line $AC$ such that $\angle PBA = \angle QBA = \angle ACB $ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $BQ$ for which $PD=PB$. Let the ray $AD$ intersect the circle $ABC$ at $R \neq A$. Prove that $QB = QR$.

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.

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 $ABC$ be a triangle with $\angle B > \angle C$. Let $P$ and $Q$ be two different points on line $AC$ such that $\angle PBA = \angle QBA = \angle ACB $ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $BQ$ for which $PD=PB$. Let the ray $AD$ intersect the circle $ABC$ at $R \neq A$. Prove that $QB = QR$.

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 $P$ 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)

The inscribed circle of the triangle $ABC$ touches its sides $AB, BC, CA$, at points $K,N, M$ respectively. It is known that $\angle ANM = \angle CKM$. Prove that the triangle $ABC$ is isosceles. (Vyacheslav Yasinsky)

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.

Find all isosceles triangles that cannot be cut into three isosceles triangles with the same sides.

Show that any triangle can be subdivided into isosceles triangles.

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 $ABC$ 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]

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 $BD$.

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 $ABCD$ 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]

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, 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.

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 $\angle ACB = 90^{\circ}$.

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]

A point $D$ lie on the lateral side $BC$ of an isosceles triangle $ABC$. The ray $AD$ meets the line passing through $B$ and parallel to the base $AC$ at point $E$. Prove that the tangent to the circumcircle of triangle $ABD$ at $B$ bisects $EC$.

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]