In the acute-angled triangle $ABC$ on the sides $AC$ and $BC$, points $D$ and $E$ are chosen such that points $A, B, E$, and $D$ lie on one circle. The circumcircle of triangle $DEC$ intersects side $AB$ at points $X$ and $Y$. Prove that the midpoint of segment $XY$ is the foot of the altitude of the triangle, drawn from point $C$.

The incircle $\omega$ of triangle $ABC$ touches its sides $BC, CA$ and $AB$ at points $D, E$ and $E$, respectively. Point $G$ lies on circle $\omega$ in such a way that $FG$ is a diameter. Lines $EG$ and $FD$ intersect at point $H$. Prove that $AB \parallel CH$.

The pentagon $ABCDE$ is inscribed in the circle $\omega$. Let $T$ be the intersection point of the diagonals $BE$ and $AD$. A line is drawn through the point $T$ parallel to $CD$, which intersects $AB$ and $CE$ at points $X$ and $Y$, respectively. Prove that the circumscribed circle of the triangle $AXY$ is tangent to $\omega$.

In the triangle $ABC$, the heights $AA_1$ and $BB_1$ are drawn. On the side $AB$, points $M$ and $K$ are chosen so that $B_1K\parallel BC$ and $A_1 M\parallel AC$. Prove that the angle $AA_1K$ is equal to the angle $BB_1M$.

Let $ABCD$ be a square. The points $N$ and $P$ are chosen on the sides $AB$ and $AD$ respectively, such that $NP=NC$. The point $Q$ on the segment $AN$ is such that that $\angle QPN=\angle NCB$. Prove that $\angle BCQ=\frac{1}{2}\angle AQP$.

In the convex quadrilateral of the $ABCD$, the diagonals of $AC$ and $BD$ are mutually perpendicular and intersect at point $E$. On the side of $AD$, a point $P$ is chosen such that $PE = EC$. The circumscribed circle of the triangle $BCD$ intersects the segment $AD$ at the point $Q$. The circle passing through point $A$ and tangent to the line $EP$ at point $P$ intersects the segment $AC$ at point $R$. It turns out that points $B, Q, R$ are collinear. Prove that $\angle BCD = 90^o$.

On side $AB$ of triangle $ABC$, point $M$ is selected. A straight line passing through $M$ intersects the segment $AC$ at point $N$ and the ray $CB$ at point $K$. The circumscribed circle of the triangle $AMN$ intersects $\omega$, the circumscribed circle of the triangle $ABC$, at points $A$ and $S$. Straight lines $SM$ and $SK$ intersect with $\omega$ for the second time at points $P$ and $Q$, respectively. Prove that $AC = PQ$.

In the rectangle $ABCD$, point $M$ is the midpoint of the side $BC$. The points $P$ and $Q$ lie on the diagonal $AC$ such that $\angle DPC = \angle DQM = 90^o$. Prove that $Q$ is the midpoint of the segment $AP$.

Let $AD$ be the bisector of an acute-angled triangle $ABC$. The circle circumscribed around the triangle $ABD$ intersects the straight line perpendicular to $AD$ that passes through point $B$, at point $E$. Point $O$ is the center of the circumscribed circle of triangle $ABC$. Prove that the points $A, O, E$ lie on the same line.

The circle $\omega$ passes through the vertices $B$ and $C$ of triangle $ABC$ and intersects its sides $AC,AB$ at points $A,E$, respectively. On the ray $BD$, a point $K$ such that $BK = AC$ is chosen , and on the ray $CE$, a point $L$ such that $CL = AB$ is chosen. Prove that the center $O$ of the circumscribed circle of the triangle $AKL$ lies on the circle $\omega$.

In the quadrangle $ABCD$, the angle at the vertex $A$ is right. Point $M$ is the midpoint of the side $BC$. It turned out that $\angle ADC = \angle BAM$. Prove that $\angle ADB = \angle CAM$.

The quadrilateral $ABCD$ is inscribed in the circle $\omega$. Lines $AD$ and $BC$ intersect at point $E$. Points $M$ and $N$ are selected on segments $AD$ and $BC$, respectively, so that $AM: MD = BN: NC$. The circumscribed circle of the triangle $EMN$ intersects the circle $\omega$ at points $X$ and $Y$. Prove that the lines $AB, CD$ and $XY$ intersect at the same point or are parallel.

On the sides $AB, AC ,BC$ of the triangle $ABC$, the points $M, N, K$ are selected, respectively, such that $AM = AN$ and $BM = BK$. The circle circumscribed around the triangle $MNK$ intersects the segments $AB$ and $BC$ for the second time at points $P$ and $Q$, respectively. Lines $MN$ and $PQ$ intersect at point $T$. Prove that the line $CT$ bisects the segment $MP$.

The line $\ell$ parallel to the side $BC$ of the triangle $ABC$, intersects its sides $AB,AC$ at the points $D,E$, respectively. The circumscribed circle of triangle $ABC$ intersects line $\ell$ at points $F$ and $G$, such that points $F,D,E,G$ lie on line $\ell$ in this order. The circumscribed circles of the triangles $FEB$ and $DGC$ intersect at points $P$ and $Q$. Prove that points $A, P$ and $Q$ are collinear.

In triangle $ABC$, point$ I$ is incenter , $I_a$ is the $A$-excenter. Let $K$ be the intersection point of the $BC$ with the external bisector of the angle $BAC$, and $E$ be the midpoint of the arc $BAC$ of the circumcircle of triangle $ABC$. Prove that $K$ is the orthocenter of triangle $II_aE$.

In the acute-angled triangle $ABC$, let $CD, AE$ be the altitudes. Points $F$ and $G$ are the projections of $A$ and $C$ on the line $DE$, respectively, $H$ and $K$ are the projections of $D$ and $E$ on the line $AC$, respectively. The lines $HF$ and $KG$ intersect at point $P$. Prove that line $BP$ bisects the segment $DE$.

The inscribed circle $\Omega$ of triangle $ABC$ touches the sides $AB$ and $AC$ at points $K$ and $ L$, respectively. The line $BL$ intersects the circle $\Omega$ for the second time at the point $M$. The circle $\omega$ passes through the point $M$ and is tangent to the lines $AB$ and $BC$ at the points $P$ and $Q$, respectively. Let $N$ be the second intersection point of circles $\omega$ and $\Omega$, which is different from $M$. Prove that if $KM \parallel AC$ then the points $P, N$ and $L$ lie on one line.

In the triangle $ABC$, the segment $CL$ is the angle bisector. The $C$-exscribed circle with center at the point $ I_c$ touches the side of the $AB$ at the point $D$ and the extension of sides $CA$ and $CB$ at points $P$ and $Q$, respectively. It turned out that the length of the segment $CD$ is equal to the radius of this exscribed circle. Prove that the line $PQ$ bisects the segment $I_CL$.