Let $ABCD$ be a parallelogram, $AC$ intersects $BD$ at $I$. Consider point $G$ inside $\triangle ABC$ that satisfy $\angle IAG=\angle IBG\neq 45^{\circ}-\frac{\angle AIB}{4}$. Let $E,G$ be projections of $C$ on $AG$ and $D$ on $BG$. The $E-$ median line of $\triangle BEF$ and $F-$ median line of $\triangle AEF$ intersects at $H$. $a)$ Prove that $AF,BE$ and $IH$ concurrent. Call the concurrent point $L$. $b)$ Let $K$ be the intersection of $CE$ and $DF$. Let $J$ circumcenter of $(LAB)$ and $M,N$ are respectively be circumcenters of $(EIJ)$ and $(FIJ)$. Prove that $EM,FN$ and the line go through circumcenters of $(GAB),(KCD)$ are concurrent.

In an acute-angled triangle $ABC$ the interior bisector of angle $A$ meets $BC$ at $L$ and meets the circumcircle of $ABC$ again at $N$. From $L$ perpendiculars are drawn to $AB$ and $AC$, with feet $K$ and $M$ respectively. Prove that the quadrilateral $AKNM$ and the triangle $ABC$ have equal areas.

In triangle $\triangle ABC$, $AB<AC$, $M,D,E$ are the midpoints of $BC$, the arcs $BAC$ and $BC$ of the circumcircle of $\triangle ABC$ respectively. The incircle of $\triangle ABC$ touches $AB$ at $F$, $AE$ meets $BC$ at $G$, and the perpendicular to $AB$ at $B$ meets segment $EF$ at $N$. If $BN=EM$, prove that $DF$ is perpendicular to $FG$.

Consider an acute-angled triangle $ABC$ with $AB<AC$ and let $\omega$ be its circumscribed circle. Let $t_B$ and $t_C$ be the tangents to the circle $\omega$ at points $B$ and $C$, respectively, and let $L$ be their intersection. The straight line passing through the point $B$ and parallel to $AC$ intersects $t_C$ in point $D$. The straight line passing through the point $C$ and parallel to $AB$ intersects $t_B$ in point $E$. The circumcircle of the triangle $BDC$ intersects $AC$ in $T$, where $T$ is located between $A$ and $C$. The circumcircle of the triangle $BEC$ intersects the line $AB$ (or its extension) in $S$, where $B$ is located between $S$ and $A$. Prove that $ST$, $AL$, and $BC$ are concurrent. $\text{Vangelis Psychas and Silouanos Brazitikos}$

Let $ABCD$ be a cyclic quadrilateral inscirbed in a circle $O$ ($AB> AD$), and let $E$ be a point on segment $AB$ such that $AE = AD$. Let $AC \cap DE = F$, and $DE \cap O = K(\ne D)$. The tangent to the circle passing through $C,F,E$ at $E$ hits $AK$ at $L$. Prove that $AL = AD$ if and only if $\angle KCE = \angle ALE$.

Let $ABCD$ be a parallelogram with $AC=BC.$ A point $P$ is chosen on the extension of ray $AB$ past $B.$ The circumcircle of $ACD$ meets the segment $PD$ again at $Q.$ The circumcircle of triangle $APQ$ meets the segment $PC$ at $R.$ Prove that lines $CD,AQ,BR$ are concurrent.

Let $ABCD$ be a cyclic quadrilateral with $|AB|=26$, $|BC|=10$, $m(\widehat{ABD})=45^\circ$,$m(\widehat{ACB})=90^\circ$. What is the area of $\triangle DAC$ ? $ \textbf{(A)}\ 120 \qquad\textbf{(B)}\ 108 \qquad\textbf{(C)}\ 90 \qquad\textbf{(D)}\ 84 \qquad\textbf{(E)}\ 80 $

Let $ABC$ be a triangle such that $AB > BC$ and let $D$ be a variable point on the line segment $BC$. Let $E$ be the point on the circumcircle of triangle $ABC$, lying on the opposite side of $BC$ from $A$ such that $\angle BAE = \angle DAC$. Let $I$ be the incenter of triangle $ABD$ and let $J$ be the incenter of triangle $ACE$. Prove that the line $IJ$ passes through a fixed point, that is independent of $D$. [i]Proposed by Merlijn Staps[/i]

Denote by $ M$ midpoint of side $ BC$ in an isosceles triangle $ \triangle ABC$ with $ AC = AB$. Take a point $ X$ on a smaller arc $ \overarc{MA}$ of circumcircle of triangle $ \triangle ABM$. Denote by $ T$ point inside of angle $ BMA$ such that $ \angle TMX = 90$ and $ TX = BX$. Prove that $ \angle MTB - \angle CTM$ does not depend on choice of $ X$. [i]Author: Farzan Barekat, Canada[/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$.

The diagonals of a trapezoid $ ABCD$ intersect at point $ P$. Point $ Q$ lies between the parallel lines $ BC$ and $ AD$ such that $ \angle AQD \equal{} \angle CQB$, and line $ CD$ separates points $ P$ and $ Q$. Prove that $ \angle BQP \equal{} \angle DAQ$. [i]Author: Vyacheslav Yasinskiy, Ukraine[/i]

Let $ABCD$ be a cyclic quadrilateral whose sides have pairwise different lengths. Let $O$ be the circumcenter of $ABCD$. The internal angle bisectors of $\angle ABC$ and $\angle ADC$ meet $AC$ at $B_1$ and $D_1$, respectively. Let $O_B$ be the center of the circle which passes through $B$ and is tangent to $\overline{AC}$ at $D_1$. Similarly, let $O_D$ be the center of the circle which passes through $D$ and is tangent to $\overline{AC}$ at $B_1$. Assume that $\overline{BD_1} \parallel \overline{DB_1}$. Prove that $O$ lies on the line $\overline{O_BO_D}$.

Triangle $ABC$ has circumcenter $O$. $D$ is an arbitrary point on $BC$, and $AD$ intersects $\odot(ABC)$ at $E$. $S$ is a point on $\odot(ABC)$ such that $D, O, E, S$ are colinear. $AS$ intersects $BC$ at $P$. $Q$ is a point on $BC$ such that $D, O, A, Q$ are concylic. Prove that $\odot(ABC)$ is tangent to $\odot (APQ)$. [i]Proposed by chengbilly[/i]

Let $ABC$ be an acute triangle.The lines $l_1$ and $l_2$ are perpendicular to $AB$ at the points $A$ and $B$, respectively.The perpendicular lines from the midpoint $M$ of $AB$ to the lines $AC$ and $BC$ intersect $l_1$ and $l_2$ at the points $E$ and $F$, respectively.If $D$ is the intersection point of the lines $EF$ and $MC$, prove that \[\angle ADB = \angle EMF.\]