Let $ABC$ be a triangle such that $BC>AB$, $L$ be the internal angle bisector of $\angle ABC$. Let $P,Q$ be the feet from $A,C$ to $L$, respectively. Suppose $M,N$ are the midpoints of $\overline{AC}$ and $\overline{BC}$, respectively. Let $O$ be the circumcenter of triangle $PQM$, and the circumcircle intersects $AC$ at point $H$. Prove that $O,M,N,H$ are concyclic.

Let $ABC$ be a triangle with $\angle C = 90^\circ$ and $CA \neq CB$. Let $CH$ be an altitude and $CL$ be an interior angle bisector. Show that for $X \neq C$ on the line $CL$, we have $\angle XAC \neq \angle XBC$. Also show that for $Y \neq C$ on the line $CH$ we have $\angle YAC \neq \angle YBC$. [i]Bulgaria[/i]

Let $D$ be a point on side $[AB]$ of triangle $ABC$ with $|AB|=|AC|$ such that $[CD]$ is an angle bisector and $m(\widehat{ABC})=40^\circ$. Let $F$ be a point on the extension of $[AB]$ after $B$ such that $|BC|=|AF|$. Let $E$ be the midpoint of $[CF]$. If $G$ is the intersection of lines $ED$ and $AC$, what is $m(\widehat{FBG})$? $ \textbf{(A)}\ 150^\circ \qquad\textbf{(B)}\ 135^\circ \qquad\textbf{(C)}\ 120^\circ \qquad\textbf{(D)}\ 105^\circ \qquad\textbf{(E)}\ \text{None of above} $

On side, $BC, AB$ of a parallelogram $ABCD$ lie points $M,N$ respectively such that $|AM| =|CN|$. Let $P$ be the intersection of $AM$ and $CN$. Prove that the angle bisector of $\angle APC$ passes through $D$.

Acute-angled triangle $ABC$ is inscribed into circle $\Omega$. Lines tangent to $\Omega$ at $B$ and $C$ intersect at $P$. Points $D$ and $E$ are on $AB$ and $AC$ such that $PD$ and $PE$ are perpendicular to $AB$ and $AC$ respectively. Prove that the orthocentre of triangle $ADE$ is the midpoint of $BC$.

Given triangle $ABC$, let $M$ be the midpoint of side $AB$ and $N$ be the midpoint of side $AC$. A circle is inscribed inside quadrilateral $NMBC$, tangent to all four sides, and that circle touches $MN$ at point $X.$ The circle inscribed in triangle $AMN$ touches $MN$ at point $Y$, with $Y$ between $X$ and $N$. If $XY=1$ and $BC=12$, find, with proof, the lengths of the sides $AB$ and $AC$.

Let $ABC$ be an isosceles triangle with apex at $C.$ Let $D$ and $E$ be two points on the sides $AC$ and $BC$ such that the angle bisectors $\angle DEB$ and $\angle ADE$ meet at $F,$ which lies on segment $AB.$ Prove that $F$ is the midpoint of $AB.$

Let $BL, AD$ be the bisector and the altitude correspondingly of an acute triangle ABC. They intersect at point $T$. It turned out that the altitude $LK$ of $\triangle ALB$ is divided in half by the line $AD$. Prove that $KT \perp BL$. [i]Proposed by Mariia Rozhkova[/i]

In obtuse triangle $ABC$, with the obtuse angle at $A$, let $D$, $E$, $F$ be the feet of the altitudes through $A$, $B$, $C$ respectively. $DE$ is parallel to $CF$, and $DF$ is parallel to the angle bisector of $\angle BAC$. Find the angles of the triangle.

Let $\Gamma_1$ and $\Gamma_2$ be circles internally tangent at point $A$, with $\Gamma_1$ inside $\Gamma_2$. Let $BC$ be a chord of $\Gamma_2$ which is tangent to $\Gamma_1$ at point $D$. Prove that line $AD$ is the angle bisector of $\angle BAC$.

Bisectrices $AA_1$ and $BB_1$ of triangle $ABC$ meet in $I$. Segments $A_1I$ and $B_1I$ are the bases of isosceles triangles with opposite vertices $A_2$ and $B_2$ lying on line $AB$. It is known that line $CI$ bisects segment $A_2B_2$. Is it true that triangle $ABC$ is isosceles?

Triangle $ABC$ has a right angle at $C$. Let $D$ be the midpoint of side $\overline{AC}$, and let $E$ be the intersection of $\overline{AC}$ and the bisector of $\angle ABC$. The area of $\triangle ABC$ is $144$, and the area of $\triangle DBE$ is $8$. Find $AB^2$.

Triangle $ABC$ is isosceles, with $AB=AC$ and altitude $AM=11.$ Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$ Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt{b},$ where $a$ and $b$ are integers. Find $a+b.$ [asy] import graph; size(7cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-1.55,xmax=7.95,ymin=-4.41,ymax=5.3; draw((1,3)--(0,0)); draw((0,0)--(2,0)); draw((2,0)--(1,3)); draw((1,3)--(1,0)); draw((1,0.7)--(0,0)); draw((1,0.7)--(2,0)); label("$11$",(0.75,1.63),SE*lsf); dot((1,3),ds); label("$A$",(0.96,3.14),NE*lsf); dot((0,0),ds); label("$B$",(-0.15,-0.18),NE*lsf); dot((2,0),ds); label("$C$",(2.06,-0.18),NE*lsf); dot((1,0),ds); label("$M$",(0.97,-0.27),NE*lsf); dot((1,0.7),ds); label("$D$",(1.05,0.77),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]

In $\Delta ABC$, $\angle ABC=120^\circ$. The internal bisector of $\angle B$ meets $AC$ at $D$. If $BD=1$, find the smallest possible value of $4BC+AB$.

In $\vartriangle ABC$, the bisector of $\angle B$ meets $AC$ at $D$ and the bisector of $\angle C$ meets $AB$ at $E$. These bisectors intersect at $O$ and $OD = OE$. If $AD \ne AE$, prove that $\angle A = 60^o$.