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.

For each vertex of triangle $ABC$, the angle between the altitude and the bisectrix from this vertex was found. It occurred that these angle in vertices $A$ and $B$ were equal. Furthermore the angle in vertex $C$ is greater than two remaining angles. Find angle $C$ of the triangle.

Let the points $O$ and $H$ be the circumcenter and orthocenter of an acute angled triangle $ABC.$ Let $D$ be the midpoint of $BC.$ Let $E$ be the point on the angle bisector of $\angle BAC$ such that $AE\perp HE.$ Let $F$ be the point such that $AEHF$ is a rectangle. Prove that $D,E,F$ are collinear.

We consider a square $ABCD$ and a point $E$ on the segment $CD$. The bisector of $\angle EAB$ cuts the segment $BC$ in $F$. Prove that $BF + DE = AE$.

Cyclic quadrilateral $ABCD$ is inscribed in circle $k$ with center $O$. The angle bisector of $ABD$ intersects $AD$ and $k$ in $K,M$ respectively, and the angle bisector of $CBD$ intersects $CD$ and $k$ in $L,N$ respectively. If $KL\parallel MN$ prove that the circumcircle of triangle $MON$ bisects segment $BD$.

Let $ABC$ be a triangle with $AB=16$ and $AC=5$. Suppose that the bisectors of angle $\angle ABC$ and $\angle BCA$ meet at a point $P$ in the triangle's interior. Given that $AP=4$, compute $BC$.

Let $ABC$ be a triangle. The internal bisector of $\angle B$ meets $AC$ in $P$ and $I$ is the incenter of $ABC$. Prove that if $AP+AB = CB$, then $API$ is an isosceles triangle.

An acute triangle $ ABC$ with $ AB \neq AC$ is given. Let $ V$ and $ D$ be the feet of the altitude and angle bisector from $ A$, and let $ E$ and $ F$ be the intersection points of the circumcircle of $ \triangle AVD$ with sides $ AC$ and $ AB$, respectively. Prove that $ AD$, $ BE$ and $ CF$ have a common point.

Let $ABC$ be a triangle such that length of internal angle bisector from $B$ is equal to $s$. Also, length of external angle bisector from $B$ is equal to $s_1$. Find area of triangle $ABC$ if $\frac{AB}{BC} = k$

Let $P$ be a point inside a triangle $ABC$ with $\angle C = 90^o$ such that $AP = AC$, and let $M$ be the midpoint of $AB$ and $CH$ be the altitude. Prove that $PM$ bisects $\angle BPH$ if and only if $\angle A = 60^o$.

Triangle $ ABC$ has a right angle at $ B$, $ AB \equal{} 1$, and $ BC \equal{} 2$. The bisector of $ \angle BAC$ meets $ \overline{BC}$ at $ D$. What is $ BD$? [asy]unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4; pair A=(0,1), B=(0,0), C=(2,0); pair D=extension(A,bisectorpoint(B,A,C),B,C); pair[] ds={A,B,C,D}; dot(ds); draw(A--B--C--A--D); label("$1$",midpoint(A--B),W); label("$B$",B,SW); label("$D$",D,S); label("$C$",C,SE); label("$A$",A,NW); draw(rightanglemark(C,B,A,2));[/asy]$ \textbf{(A)}\ \frac {\sqrt3 \minus{} 1}{2} \qquad \textbf{(B)}\ \frac {\sqrt5 \minus{} 1}{2} \qquad \textbf{(C)}\ \frac {\sqrt5 \plus{} 1}{2} \qquad \textbf{(D)}\ \frac {\sqrt6 \plus{} \sqrt2}{2}$ $ \textbf{(E)}\ 2\sqrt3 \minus{} 1$

Suppose $n>2$ and let $A_1,\dots,A_n$ be points on the plane such that no three are collinear. [b](a)[/b] Suppose $M_1,\dots,M_n$ be points on segments $A_1A_2,A_2A_3,\dots ,A_nA_1$ respectively. Prove that if $B_1,\dots,B_n$ are points in triangles $M_2A_2M_1,M_3A_3M_2,\dots ,M_1A_1M_n$ respectively then \[|B_1B_2|+|B_2B_3|+\dots+|B_nB_1| \leq |A_1A_2|+|A_2A_3|+\dots+|A_nA_1|\] Where $|XY|$ means the length of line segment between $X$ and $Y$. [b](b)[/b] If $X$, $Y$ and $Z$ are three points on the plane then by $H_{XYZ}$ we mean the half-plane that it's boundary is the exterior angle bisector of angle $\hat{XYZ}$ and doesn't contain $X$ and $Z$ ,having $Y$ crossed out. Prove that if $C_1,\dots ,C_n$ are points in ${H_{A_nA_1A_2},H_{A_1A_2A_3},\dots,H_{A_{n-1}A_nA_1}}$ then \[|A_1A_2|+|A_2A_3|+\dots +|A_nA_1| \leq |C_1C_2|+|C_2C_3|+\dots+|C_nC_1|\] Time allowed for this problem was 2 hours.

$ ABC$ is a triangle with $ \angle BAC \equal{} 10{}^\circ$, $ \angle ABC \equal{} 150{}^\circ$. Let $ X$ be a point on $ \left[AC\right]$ such that $ \left|AX\right| \equal{} \left|BC\right|$. Find $ \angle BXC$. $\textbf{(A)}\ 15^\circ \qquad\textbf{(B)}\ 20^\circ \qquad\textbf{(C)}\ 25^\circ \qquad\textbf{(D)}\ 30^\circ \qquad\textbf{(E)}\ 35^\circ$

Prove that the bisectors of the angles of a rectangle, extended to their mutual intersection, form a square.

Let $ABCD$ be a square, and $C$ the circle whose diameter is $AB.$ Let $Q$ be an arbitrary point on the segment $CD.$ We know that $QA$ meets $C$ on $E$ and $QB$ meets it on $F.$ Also $CF$ and $DE$ intersect in $M.$ show that $M$ belongs to $C.$

If $BK$ is an angle bisector of $\angle ABC$ in triangle $ABC$. Find angles of triangle $ABC$ if $BK=KC=2AK$

Let $ABC$ be an acute angled triangle (with $AB<AC<BC$) inscribed in circle $c(O,R)$ (with center $O$ and radius $R$). Circle $c_1(A,AB)$ (with center $A$ and radius $AB$) intersects side $BC$ at point $D$ and the circumcircle $c(O,R)$ at point $E$. Prove that side $AC$ bisects angle $\angle DAE$.

Let an acute-angled triangle $ABC$ with $AB<AC<BC$, inscribed in circle $c(O,R)$. The angle bisector $AD$ meets $c(O,R)$ at $K$. The circle $c_1(O_1,R_1)$(which passes from $A,D$ and has its center $O_1$ on $OA$) meets $AB$ at $E$ and $AC$ at $Z$. If $M,N$ are the midpoints of $ZC$ and $BE$ respectively, prove that: [b]a)[/b]the lines $ZE,DM,KC$ are concurrent at one point $T$. [b]b)[/b]the lines $ZE,DN,KB$ are concurrent at one point $X$. [b]c)[/b]$OK$ is the perpendicular bisector of $TX$.

The extension of the bisector of angle $A$ of triangle $ABC$ intersects with the circumscribed circle of this triangle at point $W$. A straight line is drawn through $W$, which is parallel to side $AB$ and intersects sides $BC$ and $AC$ , at points $N$ and $K$, respectively. Prove that the line $AW$ is tangent to the circumscribed circle of $\vartriangle CNW$. (Sergey Yakovlev)

Let $ABCD$ be a parallelogram, such that the point $M$ is the midpoint of the side $CD$ and lies on the bisector of the angle $\angle BAD$. Prove that $\angle AMB = 90^o$.

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 $AB.$

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]

On the sides $BC$ and $AC$ of the isosceles triangle $ABC$ ($AB = BC$), points $E$ and $D$ are marked, respectively, so that $DE \parallel AB$. On the extendsion of side $CB$ beyond the point $B$, point $K$ was arbitrarily marked. Let $P$ be the intersection point of the lines $AB$ and $KD$. Let $Q$ be the intersection point of the lines $AK$ and $DE$. Prove that $CA$ is the bisector of angle $\angle PCQ$.

Let $ABC$ be a triangle in which $\measuredangle{A}=135^{\circ}$. The perpendicular to the line $AB$ erected at $A$ intersects the side $BC$ at $D$, and the angle bisector of $\angle B$ intersects the side $AC$ at $E$. Find the measure of $\measuredangle{BED}$.

Given $\triangle ABC$ with circumcircle $\ell$. Point $M$ in $\triangle ABC$ such that $AM$ is the angle bisector of $\angle BAC$. Circle with center $M$ and radius $MB$ intersects $\ell$ and $BC$ at $D$ and $E$ respectively, $(B \not= D, B \not= E)$. Let $P$ be the midpoint of arc $BC$ in $\ell$ that didn't have $A$. Prove that $AP$ angle bisector of $\angle DPE$ if and only if $\angle B = 90^{\circ}$.