Cyclic quadrilateral $ABCD$ has circumcircle $(O)$. Points $M$ and $N$ are the midpoints of $BC$ and $CD$, and $E$ and $F$ lie on $AB$ and $AD$ respectively such that $EF$ passes through $O$ and $EO=OF$. Let $EN$ meet $FM$ at $P$. Denote $S$ as the circumcenter of $\triangle PEF$. Line $PO$ intersects $AD$ and $BA$ at $Q$ and $R$ respectively. Suppose $OSPC$ is a parallelogram. Prove that $AQ=AR$.

Let $M$ be the midpoint of the side $AC$ of a triangle $ABC$, and let $K$ be a point on the ray $BA$ beyond $A$. The line $KM$ intersects the side $BC$ at the point $L$. $P$is the point on the segment $BM$ such that $PM$ is the bisector of the angle $LPK$. The line $\ell$ passes through $A$ and is parallel to $BM$. Prove that the projection of the point $M$ onto the line $\ell$ belongs to the line $PK$.

On the bisector of the angle $ BAC $ of the triangle $ ABC $ we choose the points $ {{B} _ {1}}, \, \, {{C} _ {1}} $ for which $ B {{B} _ {1 }}\perp AB $, $ C {{C} _ {1}} \perp AC $. The point $ M $ is the midpoint of the segment $ {{B} _ {1}} {{C} _ {1}} $. Prove that $ MB = MC $.

Let $ABC$ be an acute-angled triangle with circumference $\Omega$. Let the angle bisectors of $\angle B$ and $\angle C$ intersect $\Omega$ again at $M$ and $N$. Let $I$ be the intersection point of these angle bisectors. Let $M'$ and $N'$ be the respective reflections of $M$ and $N$ in $AC$ and $AB$. Prove that the center of the circle passing through $I$, $M'$, $N'$ lies on the altitude of triangle $ABC$ from $A$. [i]Proposed by Victor Domínguez and Ariel García[/i]

Let $\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $ABC$ with $AB > BC$. The angle bisector of $\angle ABC$ intersects $\Omega$ at $M \ne B$. Let $\Gamma$ be the circle with diameter $BM$. The angle bisectors of $\angle AOB$ and $\angle BOC$ intersect $\Gamma$ at points $P$ and $Q,$ respectively. The point $R$ is chosen on the line $P Q$ so that $BR = MR$. Prove that $BR\parallel AC$. (Here we always assume that an angle bisector is a ray.) [i]Proposed by Sergey Berlov, Russia[/i]

[b]5.[/b] In triangle $ABC$, let $r_A$ be the line that passes through the midpoint of $BC$ and is perpendicular to the internal bisector of $\angle{BAC}$. Define $r_B$ and $r_C$ similarly. Let $H$ and $I$ be the orthocenter and incenter of $ABC$, respectively. Suppose that the three lines $r_A$, $r_B$, $r_C$ define a triangle. Prove that the circumcenter of this triangle is the midpoint of $HI$.

Let $ABC$ be a triangle such that $|AB|=8$ and $|AC|=2|BC|$. What is the largest value of altitude from side $[AB]$? $ \textbf{(A)}\ 3\sqrt 2 \qquad\textbf{(B)}\ 3\sqrt 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ \dfrac {16}3 \qquad\textbf{(E)}\ 6 $

Let $ABC$ be an acute triangle with $| AB | > | AC |$. Let $D$ be the foot of the altitude from $A$ to $BC$, let $K$ be the intersection of $AD$ with the internal bisector of angle $B$, Let $M$ be the foot of the perpendicular from $B$ to $CK$ (it could be in the extension of segment $CK$) and$ N$ the intersection of $BM$ and $AK$ (it could be in the extension of the segments). Let $T$ be the intersection of$ AC$ with the line that passes through $N$ and parallel to $DM$. Prove that $BM$ is the internal bisector of the angle $\angle TBC$

Let $ABC$ be a triangle. For any point $X$ on $BC$, let $AX$ meet the circumcircle of $ABC$ in $X'$. Prove or disprove: $XX'$ has maximum length if and only if $AX$ lies between the median and the internal angle bisector from $A$.

Circle $k$ and its diameter $AB$ are given. Find the locus of the centers of circles inscribed in the triangles having one vertex on $AB$ and two other vertices on $k.$

Let $ABC$ be a right triangle with $\angle A=90^\circ.$ The bisectors of the angles $B$ and $C$ meet each other in $I$ and meet the sides $AC$ and $AB$ in $D$ and $E$, respectively. Prove that $S_{BCDE}=2S_{BIC},$ where $S$ is the area function. [asy] import graph; size(200); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen ttqqcc = rgb(0.2,0,0.8); pen qqwuqq = rgb(0,0.39,0); pen xdxdff = rgb(0.49,0.49,1); pen fftttt = rgb(1,0.2,0.2); pen ccccff = rgb(0.8,0.8,1); draw((1.89,4.08)--(1.89,4.55)--(1.42,4.55)--(1.42,4.08)--cycle,qqwuqq); draw((1.42,4.08)--(7.42,4.1),ttqqcc+linewidth(1.6pt)); draw((1.4,10.08)--(1.42,4.08),ttqqcc+linewidth(1.6pt)); draw((1.4,10.08)--(7.42,4.1),ttqqcc+linewidth(1.6pt)); draw((1.4,10.08)--(4,4.09),fftttt+linewidth(1.2pt)); draw((7.42,4.1)--(1.41,6.24),fftttt+linewidth(1.2pt)); draw((1.41,6.24)--(4,4.09),ccccff+linetype("5pt 5pt")); dot((1.42,4.08),ds); label("$A$", (1.1,3.66),NE*lsf); dot((7.42,4.1),ds); label("$B$", (7.15,3.75),NE*lsf); dot((1.4,10.08),ds); label("$C$", (1.49,10.22),NE*lsf); dot((4,4.09),ds); label("$E$", (3.96,3.46),NE*lsf); dot((1.41,6.24),ds); label("$D$", (0.9,6.17),NE*lsf); dot((3.37,5.54),ds); label("$I$", (3.45,5.69),NE*lsf); clip((-6.47,-7.49)--(-6.47,11.47)--(16.06,11.47)--(16.06,-7.49)--cycle); [/asy]

Let $ABC$ be an equilateral triangle. $M$ is the midpoint of segment $AB$ and $N$ is the midpoint of segment $BC$. Let $P$ be the point outside $ABC$ such that the triangle $ACP$ is isosceles and right in $P$. $PM$ and $AN$ are cut in $I$. Prove that $CI$ is the bisector of the angle $MCA$ .

Let $ABC$ be a triangle and $\Gamma$ be the circle with diameter $[AB]$. The bisectors of $\angle BAC$ and $\angle ABC$ cut the circle $\Gamma$ again at $D$ and $E$, respectively. The incicrcle of the triangle $ABC$ cuts the lines $BC$ and $AC$ in $F$ and $G$, respectively. Show that the points $D, E, F$ and $G$ lie on the same line.

In the acute-angled non-isosceles triangle $ABC$, the height $AH$ is drawn. Points $B_1$ and $C_1$ are marked on the sides $AC$ and $AB$, respectively, so that $HA$ is the angle bisector of $B_1HC_1$ and quadrangle $BC_1B_1C$ is cyclic. Prove that $B_1$ and $C_1$ are feet of the altitudes of triangle $ABC$.

The angle bisector of $A$ in triangle $ABC$ intersects $BC$ in the point $D$. The point $E$ lies on the side $AC$, and the lines $AD$ and $BE$ intersect in the point $F$. Furthermore, $\frac{|AF|}{|F D|}= 3$ and $\frac{|BF|}{|F E|}=\frac{5}{3}$. Prove that $|AB| = |AC|$. [img]https://1.bp.blogspot.com/-evofDCeJWPY/XzT9dmxXzVI/AAAAAAAAMVY/ZN87X3Cg8iMiULwvMhgFrXbdd_f1f-JWwCLcBGAsYHQ/s0/2013%2BMohr%2Bp5.png[/img]

On the plane, given an angle $ xOy$. $ M$ be a mobile point on ray $ Ox$ and $ N$ a mobile point on ray $ Oy$. Let $ d$ be the external angle bisector of angle $ xOy$ and $ I$ be the intersection of $ d$ with the perpendicular bisector of $ MN$. Let $ P$, $ Q$ be two points lie on $ d$ such that $ IP \equal{} IQ \equal{} IM \equal{} IN$, and let $ K$ the intersection of $ MQ$ and $ NP$. $ 1.$ Prove that $ K$ always lie on a fixed line. $ 2.$ Let $ d_1$ line perpendicular to $ IM$ at $ M$ and $ d_2$ line perpendicular to $ IN$ at $ N$. Assume that there exist the intersections $ E$, $ F$ of $ d_1$, $ d_2$ from $ d$. Prove that $ EN$, $ FM$ and $ OK$ are concurrent.

Let $ABCD$ be a convex quadrilateral with $AB = BD = 8$ and $CD = DA = 6$. Let $P$ be a point on side $AB$ such that $DP$ is bisector of angle $\angle ADB$ and let $Q$ be a point on side $BC$ such that $DQ$ is bisector of angle $\angle CDB$. Calculate the radius of the circumcircle of triangle $DPQ$. Note: The circumcircle of a triangle is the circle that passes through its three vertices.

In $\triangle ABC$, $AB<AC$. $D$ and $P$ are the feet of the internal and external angle bisectors of $\angle BAC$, respectively. $M$ is the midpoint of segment $BC$, and $\omega$ is the circumcircle of $\triangle APD$. Suppose $Q$ is on the minor arc $AD$ of $\omega$ such that $MQ$ is tangent to $\omega$. $QB$ meets $\omega$ again at $R$, and the line through $R$ perpendicular to $BC$ meets $PQ$ at $S$. Prove $SD$ is tangent to the circumcircle of $\triangle QDM$. [i]Proposed by Ray Li[/i]

In a triangle $ABC$, let $D$ be the midpoint of the side $BC$, and let $E$ be a point on the side $AC$. The lines $BE$ and $AD$ meet at a point $F$. Prove: If $\frac{BF}{FE}=\frac{BC}{AB}+1$, then the line $BE$ bisects the angle $ABC$.

In triangle $ABC$, angle $B$ is obtuse and equal to $a$. The bisectors of angles $A$ and $C$ intersect opposite sides at points $P$ and $M$, respectively. On the side $AC$, points $K$ and $L$ are taken so that $\angle ABK = \angle CBL = 2a - 180^o$. What is the angle between straight lines $KP$ and $LM$?

In a non equilateral triangle $ABC$ the sides $a,b,c$ form an arithmetic progression. Let $I$ be the incentre and $O$ the circumcentre of the triangle $ABC$. Prove that (1) $IO$ is perpendicular to $BI$; (2) If $BI$ meets $AC$ in $K$, and $D$, $E$ are the midpoints of $BC$, $BA$ respectively then $I$ is the circumcentre of triangle $DKE$.

Let $\triangle ABC$ be a triangle with $AB < AC$. Let the angle bisector of $\angle BAC$ meet $BC$ at $D$ , and let $M$ be the midpoint of $BC$ . Let $P$ be the foot of the perpendicular from $B$ to $AD$ . $Q$ the intersection of $BP$ and $AM$ . Show that : $(DQ) // (AB) $ .

$O$ is a circumcircle of non-isosceles triangle $ABC$ and the angle bisector of $A$ meets $BC$ at $D$. If the line perpendicular to $BC$ passing through $D$ meets $AO$ at $E$, show that $ADE$ is an isosceles triangle.

In triangle $ABC$, the bisector $CL$ was drawn. The incircles of triangles $CAL$ and $CBL$ touch $AB$ at points $M$ and $N$ respectively. Points $M$ and $N$ are marked on the picture, and then the whole picture except the points $A, L, M$, and $N$ is erased. Restore the triangle using a compass and a ruler. (V.Protasov)

Let $\triangle ABC$ have incenter $I$ and centroid $G$. Suppose that $P_A$ is the foot of the perpendicular from $C$ to the exterior angle bisector of $B$, and $Q_A$ is the foot of the perpendicular from $B$ to the exterior angle bisector of $C$. Define $P_B$, $P_C$, $Q_B$, and $Q_C$ similarly. Show that $P_A, P_B, P_C, Q_A, Q_B,$ and $Q_C$ lie on a circle whose center is on line $IG$.