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

In an acute-angled triangle $ABC$, $CF$ is an altitude, with $F$ on $AB$, and $BM$ is a median, with $M$ on $CA$. Given that $BM=CF$ and $\angle MBC=\angle FCA$, prove that triangle $ABC$ is equilateral.

Let $ABC$ be a triangle such that $ABC$ isn't a isosceles triangle. $(I)$ is incircle of triangle touches $BC,CA,AB$ at $D,E,F$ respectively. The line through $E$ perpendicular to $BI$ cuts $(I)$ again at $K$. The line through $F$ perpendicular to $CI$ cuts $(I)$ again at $L$.$J$ is midpoint of $KL$. [b]a)[/b] Prove that $D,I,J$ collinear. [b]b)[/b] $B,C$ are fixed points,$A$ is moved point such that $\frac{AB}{AC}=k$ with $k$ is constant.$IE,IF$ cut $(I)$ again at $M,N$ respectively.$MN$ cuts $IB,IC$ at $P,Q$ respectively. Prove that bisector perpendicular of $PQ$ through a fixed point.

Given a non-isosceles triangle $ABC$, let $D,E$, and $F$ denote the midpoints of the sides $BC,CA$, and $AB$ respectively. The circle $BCF$ and the line $BE$ meet again at $P$, and the circle $ABE$ and the line $AD$ meet again at $Q$. Finally, the lines $DP$ and $FQ$ meet at $R$. Prove that the centroid $G$ of the triangle $ABC$ lies on the circle $PQR$. [i](United Kingdom) David Monk[/i]

Let $ABCD$ be a convex quadrilateral with $AB$ is not parallel to $CD$. Circle $\omega_1$ with center $O_1$ passes through $A$ and $B$, and touches segment $CD$ at $P$. Circle $\omega_2$ with center $O_2$ passes through $C$ and $D$, and touches segment $AB$ at $Q$. Let $E$ and $F$ be the intersection of circles $\omega_1$ and $\omega_2$. Prove that $EF$ bisects segment $PQ$ if and only if $BC$ is parallel to $AD$.

Let $ABCD$ be a cyclic quadrilateral with circumcircle $\odot O.$ The lines tangent to $\odot O$ at $A,B$ intersect at $L.$ $M$ is the midpoint of the segment $AB.$ The line passing through $D$ and parallel to $CM$ intersects $ \odot (CDL) $ at $F.$ Line $CF$ intersects $DM$ at $K,$ and intersects $\odot O$ at $E$ (different from point $C$). Prove that $EK=DK.$

Let $O$ be the circumcentre, and $\Omega$ be the circumcircle of an acute-angled triangle $ABC$. Let $P$ be an arbitrary point on $\Omega$, distinct from $A$, $B$, $C$, and their antipodes in $\Omega$. Denote the circumcentres of the triangles $AOP$, $BOP$, and $COP$ by $O_A$, $O_B$, and $O_C$, respectively. The lines $\ell_A$, $\ell_B$, $\ell_C$ perpendicular to $BC$, $CA$, and $AB$ pass through $O_A$, $O_B$, and $O_C$, respectively. Prove that the circumcircle of triangle formed by $\ell_A$, $\ell_B$, and $\ell_C$ is tangent to the line $OP$.

Let $ABC$ be a triangle with circumcircle $\omega$, incenter $I$, and $A$-excenter $I_A$. Let the incircle and the $A$-excircle hit $BC$ at $D$ and $E$, respectively, and let $M$ be the midpoint of arc $BC$ without $A$. Consider the circle tangent to $BC$ at $D$ and arc $BAC$ at $T$. If $TI$ intersects $\omega$ again at $S$, prove that $SI_A$ and $ME$ meet on $\omega$. [i]Amol Aggarwal.[/i]

Given a non-degenerate triangle $ABC$, let $A_{1}, B_{1}, C_{1}$ be the point of tangency of the incircle with the sides $BC, AC, AB$. Let $Q$ and $L$ be the intersection of the segment $AA_{1}$ with the incircle and the segment $B_{1}C_{1}$ respectively. Let $M$ be the midpoint of $B_{1}C_{1}$. Let $T$ be the point of intersection of $BC$ and $B_{1}C_{1}$. Let $P$ be the foot of the perpendicular from the point $L$ on the line $AT$. Prove that the points $A_{1}, M, Q, P$ lie on a circle.

Let $ABCD$ be a convex quadrilateral and let $P$ and $Q$ be variable points inside this quadrilateral so that $\angle APB=\angle CPD=\angle AQB=\angle CQD$. Prove that the lines $PQ$ obtained in this way all pass through a fixed point , or they are all parallel.

$AB$ is a diameter of a circle with center $O$. Let $C$ and $D$ be two different points on the circle on the same side of $AB$, and the lines tangent to the circle at points $C$ and $D$ meet at $E$. Segments $AD$ and $BC$ meet at $F$. Lines $EF$ and $AB$ meet at $M$. Prove that $E,C,M$ and $D$ are concyclic.

Let $a, b, c, d$ be positive reals strictly smaller than $1$, such that $a+b+c+d=2$. Prove that $$\sqrt{(1-a)(1-b)(1-c)(1-d)} \leq \frac{ac+bd}{2}. $$

Two circles in the plane, $\gamma_1$ and $\gamma_2$, meet at points $M$ and $N$. Let $A$ be a point on $\gamma_1$, and let $D$ be a point on $\gamma_2$. The lines $AM$ and $AN$ meet again $\gamma_2$ at points $B$ and $C$, respectively, and the lines $DM$ and $DN$ meet again $\gamma_1$ at points $E$ and $F$, respectively. Assume the order $M$, $N$, $F$, $A$, $E$ is circular around $\gamma_1$, and the segments $AB$ and $DE$ are congruent. Prove that the points $A$, $F$, $C$ and $D$ lie on a circle whose centre does not depend on the position of the points $A$ and $D$ on the respective circles, subject to the assumptions above. [i]***[/i]

Two circles $ \Omega_{1}$ and $ \Omega_{2}$ are externally tangent to each other at a point $ I$, and both of these circles are tangent to a third circle $ \Omega$ which encloses the two circles $ \Omega_{1}$ and $ \Omega_{2}$. The common tangent to the two circles $ \Omega_{1}$ and $ \Omega_{2}$ at the point $ I$ meets the circle $ \Omega$ at a point $ A$. One common tangent to the circles $ \Omega_{1}$ and $ \Omega_{2}$ which doesn't pass through $ I$ meets the circle $ \Omega$ at the points $ B$ and $ C$ such that the points $ A$ and $ I$ lie on the same side of the line $ BC$. Prove that the point $ I$ is the incenter of triangle $ ABC$. [i]Alternative formulation.[/i] Two circles touch externally at a point $ I$. The two circles lie inside a large circle and both touch it. The chord $ BC$ of the large circle touches both smaller circles (not at $ I$). The common tangent to the two smaller circles at the point $ I$ meets the large circle at a point $ A$, where the points $ A$ and $ I$ are on the same side of the chord $ BC$. Show that the point $ I$ is the incenter of triangle $ ABC$.