A quadrilateral $ABCD$ is inscribed in a circle and its two diagonals $AC,BD$ meet at $G$. Let $M$ be the center of $CD, E,F$ be the points on $BC, AD$ respectively such that $ME \parallel AC$ and $MF \parallel BD$. Point $H$ is the projection of $G$ onto $CD$. The circumcircle of $MEF$ meets $CD$ at $N$ distinct from $M$. Prove that $MN = MH$ Tran Quang Hung, Nguyen Le Phuoc, Thanh Xuan, Hanoi

An acute triangle $\triangle ABC$ has incenter $I$, and the incircle hits $BC, CA, AB$ at $D, E, F$. Lines $BI, CI, BC, DI$ hits $EF$ at $K, L, M, Q$ and the line connecting the midpoint of segment $CL$ and $M$ hits the line segment $CK$ at $P$. Prove that $$PQ=\frac{AB \cdot KQ}{BI}$$

Let $ABCD$ be a cyclic quadrilateral with $AB<CD$. The diagonals intersect at the point $F$ and lines $AD$ and $BC$ intersect at the point $E$. Let $K$ and $L$ be the orthogonal projections of $F$ onto lines $AD$ and $BC$ respectively, and let $M$, $S$ and $T$ be the midpoints of $EF$, $CF$ and $DF$ respectively. Prove that the second intersection point of the circumcircles of triangles $MKT$ and $MLS$ lies on the segment $CD$. [i](Greece - Silouanos Brazitikos)[/i]

Let $ABC$ be an acute scalene triangle with circumcenter $O$. Let $D$ be the foot of the altitude from $A$ to the side $BC$. The lines $BC$ and $AO$ intersect at $E$. Let $s$ be the line through $E$ perpendicular to $AO$. The line $s$ intersects $AB$ and $AC$ at $K$ and $L$, respectively. Denote by $\omega$ the circumcircle of triangle $AKL$. Line $AD$ intersects $\omega$ again at $X$. Prove that $\omega$ and the circumcircles of triangles $ABC$ and $DEX$ have a common point.

In an inscribed quadrilateral $ABCD$, we have $BC=CD$ but $AB\neq AD$. Points $I$ and $J$ are the incenters of triangles $ABC$ and $ACD$ respectively. Point $K$ was chosen on segment $AC$ so that $IK=JK$. Points $M$ and $N$ are the incenters of triangles $AIK$ and $AJK$. Prove that the perpendicular to $CD$ at $D$ and the perpendicular to $KI$ at $I$ intersect on the circumcircle of $MAN$.

Let $ABC$ be a triangle with $AB > AC$. The internal angle bisector of $\angle BAC$ intersects the side $BC$ at $D$. The circles with diameters $BD$ and $CD$ intersect the circumcircle of $\triangle ABC$ a second time at $P \not= B$ and $Q \not= C$, respectively. The lines $PQ$ and $BC$ intersect at $X$. Prove that $AX$ is tangent to the circumcircle of $\triangle ABC$.

An acute triangle is a triangle that has all angles less that $90^{\circ}$ ($90^{\circ}$ is a Right Angle). Let $ABC$ be a right-angled triangle with $\angle ACB =90^{\circ}.$ Let $CD$ be the altitude from $C$ to $AB,$ and let $E$ be the intersection of the angle bisector of $\angle ACD$ with $AD.$ Let $EF$ be the altitude from $E$ to $BC.$ Prove that the circumcircle of $BEF$ passes through the midpoint of $CE.$

A point $D$ is chosen on the side $AC$ of a triangle $ABC$ with $\angle C < \angle A < 90^\circ$ in such a way that $BD=BA$. The incircle of $ABC$ is tangent to $AB$ and $AC$ at points $K$ and $L$, respectively. Let $J$ be the incenter of triangle $BCD$. Prove that the line $KL$ intersects the line segment $AJ$ at its midpoint.

Triangle $ABC$ has circumcircle $\Omega$ and circumcenter $O$. A circle $\Gamma$ with center $A$ intersects the segment $BC$ at points $D$ and $E$, such that $B$, $D$, $E$, and $C$ are all different and lie on line $BC$ in this order. Let $F$ and $G$ be the points of intersection of $\Gamma$ and $\Omega$, such that $A$, $F$, $B$, $C$, and $G$ lie on $\Omega$ in this order. Let $K$ be the second point of intersection of the circumcircle of triangle $BDF$ and the segment $AB$. Let $L$ be the second point of intersection of the circumcircle of triangle $CGE$ and the segment $CA$. Suppose that the lines $FK$ and $GL$ are different and intersect at the point $X$. Prove that $X$ lies on the line $AO$. [i]Proposed by Greece[/i]

Let $ABC$ be a triangle, and let $X$, $Y$, and $Z$ be collinear points such that $AY=AZ$, $BZ=BX$, and $CX=CY$. Points $X'$, $Y'$, and $Z'$ are the reflections of $X$, $Y$, and $Z$ over $BC$, $CA$, and $AB$, respectively. Prove that if $X'Y'Z'$ is a nondegenerate triangle, then its circumcenter lies on the circumcircle of $ABC$. [i]Michael Ren[/i]

Given is an acute triangle $ABC$ and the median $AM.$ Draw $BH\perp AC.$ The line which goes through $A$ and is perpendicular to $AM$ intersects $BH$ at $E.$ On the opposite ray of the ray $AE$ choose $F$ such that $AE=AF.$ Prove that $CF\perp AB.$

Let $ABC$ be an acute triangle with circumcircle $\Gamma$. Let $\ell$ be a tangent line to $\Gamma$, and let $\ell_a, \ell_b$ and $\ell_c$ be the lines obtained by reflecting $\ell$ in the lines $BC$, $CA$ and $AB$, respectively. Show that the circumcircle of the triangle determined by the lines $\ell_a, \ell_b$ and $\ell_c$ is tangent to the circle $\Gamma$. [i]Proposed by Japan[/i]

Let $D$, $E$ and $F$ be points in which incircle of triangle $ABC$ touches sides $BC$, $CA$ and $AB$, respectively, and let $I$ be a center of that circle.Furthermore, let $P$ be a foot of perpendicular from point $I$ to line $AD$, and let $M$ be midpoint of $DE$. If $\{N\}=PM\cap{AC}$, prove that $DN \parallel EF$

Let $ \omega $ be the incircle of the triangle $ABC$ and with centre $I$. Let $\Gamma $ be the circumcircle of the triangle $AIB$. Circles $ \omega $ and $ \Gamma $ intersect at the point $X$ and $Y$. Let $Z$ be the intersection of the common tangents of the circles $\omega$ and $\Gamma$. Show that the circumcircle of the triangle $XYZ$ is tangent to the circumcircle of the triangle $ABC$.