Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ meet at $E$. The extensions of the sides $AD$ and $BC$ beyond $A$ and $B$ meet at $F$. Let $G$ be the point such that $ECGD$ is a parallelogram, and let $H$ be the image of $E$ under reflection in $AD$. Prove that $D,H,F,G$ are concyclic.

In $\triangle ABC$ $\angle B$ is obtuse and $AB \ne BC$. Let $O$ is the circumcenter and $\omega$ is the circumcircle of this triangle. $N$ is the midpoint of arc $ABC$. The circumcircle of $\triangle BON$ intersects $AC$ on points $X$ and $Y$. Let $BX \cap \omega = P \ne B$ and $BY \cap \omega = Q \ne B$. Prove that $P, Q$ and reflection of $N$ with respect to line $AC$ are collinear.

Let $ABC$ be an acute-angled triangle with $AC>BC$ and circumcircle $\omega$. Suppose that $P$ is a point on $\omega$ such that $AP=AC$ and that $P$ is an interior point on the shorter arc $BC$ of $\omega$. Let $Q$ be the intersection point of the lines $AP$ and $BC$. Furthermore, suppose that $R$ is a point on $\omega$ such that $QA=QR$ and $R$ is an interior point of the shorter arc $AC$ of $\omega$. Finally, let $S$ be the point of intersection of the line $BC$ with the perpendicular bisector of the side $AB$. Prove that the points $P, Q, R$ and $S$ are concyclic. [i]Proposed by Patrik Bak, Slovakia[/i]

Let $AA_1$ be the bisector of a triangle $ABC$. Points $D$ and $F$ are chosen on the line $BC$ such that $A_1$ is the midpoint of the segment $DF$. A line $l$, different from $BC$, passes through $A_1$ and intersects the lines $AB$ and $AC$ at points $B_1$ and $C_1$, respectively. Find the locus of the points of intersection of the lines $B_1D$ and $C_1F$ for all possible positions of $l$. [i](M. Karpuk)[/i]

Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ meet at $E$. The extensions of the sides $AD$ and $BC$ beyond $A$ and $B$ meet at $F$. Let $G$ be the point such that $ECGD$ is a parallelogram, and let $H$ be the image of $E$ under reflection in $AD$. Prove that $D,H,F,G$ are concyclic.

The circle $\omega$ with center $O$ has given. From an arbitrary point $T$ outside of $\omega$ draw tangents $TB$ and $TC$ to it. $K$ and $H$ are on $TB$ and $TC$ respectively. [b]a)[/b] $B'$ and $C'$ are the second intersection point of $OB$ and $OC$ with $\omega$ respectively. $K'$ and $H'$ are on angle bisectors of $\angle BCO$ and $\angle CBO$ respectively such that $KK' \bot BC$ and $HH'\bot BC$. Prove that $K,H',B'$ are collinear if and only if $H,K',C'$ are collinear. [b]b)[/b] Consider there exist two circle in $TBC$ such that they are tangent two each other at $J$ and both of them are tangent to $\omega$.and one of them is tangent to $TB$ at $K$ and other one is tangent to $TC$ at $H$. Prove that two quadrilateral $BKJI$ and $CHJI$ are cyclic ($I$ is incenter of triangle $OBC$).

$AB$ and $AC$ are tangents to a circle $\omega$ with center $O$ at $B,C$ respectively. Point $P$ is a variable point on minor arc $BC$. The tangent at $P$ to $\omega$ meets $AB,AC$ at $D,E$ respectively. $AO$ meets $BP,CP$ at $U,V$ respectively. The line through $P$ perpendicular to $AB$ intersects $DV$ at $M$, and the line through $P$ perpendicular to $AC$ intersects $EU$ at $N$. Prove that as $P$ varies, $MN$ passes through a fixed point.

Let $D$ be an interior point of the acute triangle $ABC$ with $AB > AC$ so that $\angle DAB = \angle CAD.$ The point $E$ on the segment $AC$ satisfies $\angle ADE =\angle BCD,$ the point $F$ on the segment $AB$ satisfies $\angle FDA =\angle DBC,$ and the point $X$ on the line $AC$ satisfies $CX = BX.$ Let $O_1$ and $O_2$ be the circumcenters of the triangles $ADC$ and $EXD,$ respectively. Prove that the lines $BC, EF,$ and $O_1O_2$ are concurrent.

Let $ABCD$ be a trapezium, $AD\parallel BC$, and let $E,F$ be points on the sides$AB$ and $CD$, respectively. The circumcircle of $AEF$ meets $AD$ again at $A_1$, and the circumcircle of $CEF$ meets $BC$ again at $C_1$. Prove that $A_1C_1,BD,EF$ are concurrent.

In a triangle $ABC$ with $AB=AC$, $\omega$ is the circumcircle and $O$ its center. Let $D$ be a point on the extension of $BA$ beyond $A$. The circumcircle $\omega_{1}$ of triangle $OAD$ intersects the line $AC$ and the circle $\omega$ again at points $E$ and $G$, respectively. Point $H$ is such that $DAEH$ is a parallelogram. Line $EH$ meets circle $\omega_{1}$ again at point $J$. The line through $G$ perpendicular to $GB$ meets $\omega_{1}$ again at point $N$ and the line through $G$ perpendicular to $GJ$ meets $\omega$ again at point $L$. Prove that the points $L, N, H, G$ lie on a circle.