Given a triangle $ABC$, in which $AB = BC$. Point $O$ is the center of the circumcircle, point $I$ is the center of the incircle. Point $D$ lies on the side $BC$, such that the lines $DI$ and $AB$ parallel. Prove that the lines $DO$ and $CI$ are perpendicular. (Vyacheslav Yasinsky)

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 $QB = QR$.

Bisectrices $AA_1$ and $BB_1$ of triangle $ABC$ meet in $I$. Segments $A_1I$ and $B_1I$ are the bases of isosceles triangles with opposite vertices $A_2$ and $B_2$ lying on line $AB$. It is known that line $CI$ bisects segment $A_2B_2$. Is it true that triangle $ABC$ is isosceles?

Each diagonal of a quadrangle divides it into two isosceles triangles. Is it true that the quadrangle is a diamond?

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 $QB = QR$.

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)