Given an acute triangle $ABC$ and points $D$, $E$, $F$ on sides $BC$, $CA$ and $AB$, respectively. If the lines $DA$, $EB$ and $FC$ are the angle bisectors of triangle $DEF$, prove that the three lines are the altitudes of triangle $ABC$.

The incenter of $\triangle A_1A_2A_3$ is $I$, and the center of the $A_i$-excircle is $J_i$ ($i=1,2,3$). Let $B_i$ be the intersection point of side $A_{i+1}A_{i+2}$ and the bisector of $\angle A_{i+1}IA_{i+2}$ ($A_{i+3}:=A_i$ $\forall i$). Prove that the three lines $B_iJ_i$ are concurrent.

Consider a triangle $ABC$ and a point $D$ on its side $\overline{AB}$. Let $I$ be a point inside $\triangle ABC$ on the angle bisector of $ACB$. The second intersections of lines $AI$ and $CI$ with circle $ACD$ are $P$ and $Q$, respectively. Similarly, the second intersection of lines $BI$ and $CI$ with circle $BCD$ are $R$ and $S$, respectively. Show that if $P\neq Q$ and $R\neq S$, then lines $AB$, $PQ$ and $RS$ pass through a point or are parallel.

Point $X$ is chosen inside the non trapezoid quadrilateral $ABCD$ such that $\angle AXD +\angle BXC=180$. Suppose the angle bisector of $\angle ABX$ meets the $D$-altitude of triangle $ADX$ in $K$, and the angle bisector of $\angle DCX$ meets the $A$-altitude of triangle $ADX$ in $L$.We know $BK \perp CX$ and $CL \perp BX$. If the circumcenter of $ADX$ is on the line $KL$ prove that $KL \perp AD$. Proposed by [i]Alireza Dadgarnia[/i]

Let $\vartriangle ABC$ be a triangle such that angle $\angle ABC$ is less than angle $\angle ACB$. The bisector of angle $\angle BAC$ cuts side $BC$ at $D$. Let $E$ be on side $AB$ such that $\angle EDB = 90^o$ and $F$ on side $AC$ such that $\angle BED = \angle DEF$. Prove that $\angle BAD = \angle FDC$.

Let triangle $ABC$ be an isosceles triangle with $AB = AC$. Suppose that the angle bisector of its angle $\angle B$ meets the side $AC$ at a point $D$ and that $BC = BD+AD$. Determine $\angle A$.

Let $M$ and $N$ be touching points of incircle with sides $AB$ and $AC$ of triangle $ABC$, and $P$ intersection point of line $MN$ and angle bisector of $\angle ABC$. Prove that $\angle BPC =90 ^{\circ}$

Let $X$ be an arbitrary point on side $BC$ of triangle ABC. Triangle $T$ formed by the bisectors of the angles $ABC$, $ACB$ and $AXC$. Prove that: a) the circumscribed circle of the triangle $T$ passes through the vertex $A$. b) the orthocenter of triangle $T$ lies on line $BC$. (Dmytro Prokopenko)

1. Let $ABC$ be an acute-angled triangle with $AB\neq AC$. The circle with diameter $BC$ intersects the sides $AB$ and $AC$ at $M$ and $N$ respectively. Denote by $O$ the midpoint of the side $BC$. The bisectors of the angles $\angle BAC$ and $\angle MON$ intersect at $R$. Prove that the circumcircles of the triangles $BMR$ and $CNR$ have a common point lying on the side $BC$.

In triangle $ ABC$, $ AB \equal{} 13$, $ BC \equal{} 14$, and $ AC \equal{} 15$. Let $ D$ denote the midpoint of $ \overline{BC}$ and let $ E$ denote the intersection of $ \overline{BC}$ with the bisector of angle $ BAC$. Which of the following is closest to the area of the triangle $ ADE$? $ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 2.5 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 3.5 \qquad \textbf{(E)}\ 4$

Let $ABC$ be an acute triangle and $D$ be an intersection of the angle bisector of $A$ and side $BC$. Let $\Omega$ be a circle tangent to the circumcircle of triangle $ABC$ and side $BC$ at $A$ and $D$, respectively. $\Omega$ meets the sides $AB, AC$ again at $E, F$, respectively. The perpendicular line to $AD$, passing through $E, F$ meets $\Omega$ again at $G, H$, respectively. Suppose that $AE$ and $GD$ meet at $P$, $EH$ and $GF$ meet at $Q$, and $HD$ and $AF$ meet at $R$. Prove that $\dfrac{\overline{QF}}{\overline{QG}}=\dfrac{\overline{HR}}{\overline{PG}}$.

Angle bisector at $A$, altitude from $B$ to $CA$ and altitude of $C$ to $AB$ on a scalene triangle $ABC$ forms a triangle $\triangle$. Let $P$ and $Q$ points on lines $AB$ and $AC$, respectively, such that the midpoint of segment $PQ$ is the orthocenter of the triangle $\triangle$. Prove that the points $B, C, P$ and $Q$ lie on a circle.

Let $AH$, $BL$ and $CM$ be an altitude, a bisectrix and a median in triangle $ABC$. It is known that lines $AH$ and $BL$ are an altitude and a bisectrix of triangle $HLM$. Prove that line $CM$ is a median of this triangle.

Let $ ABC$ be a triangle with circum-circle $ \Gamma$. Let $ M$ be a point in the interior of triangle $ ABC$ which is also on the bisector of $ \angle A$. Let $ AM, BM, CM$ meet $ \Gamma$ in $ A_{1}, B_{1}, C_{1}$ respectively. Suppose $ P$ is the point of intersection of $ A_{1}C_{1}$ with $ AB$; and $ Q$ is the point of intersection of $ A_{1}B_{1}$ with $ AC$. Prove that $ PQ$ is parallel to $ BC$.

In triangle $ABC$, sides $CB$ and $CA$ are equal to $a$ and $b$, respectively. The bisector of the angle $ACB$ intersects the side $AB$ at the point $K$ and the circumscribed circle of the triangle at point $M$. The circumscribed circle of the triangle $AMK$ intersects for second time the straight line $CA$ at the point $P$. Find the length of the segment $AP$.

Let $ABCD$ be a trapezium (trapezoid) with $AD$ parallel to $BC$ and $J$ be the intersection of the diagonals $AC$ and $BD$. Point $P$ a chosen on the side $BC$ such that the distance from $C$ to the line $AP$ is equal to the distance from $B$ to the line $DP$. [i]The following three questions 1, 2 and 3 are independent, so that a condition in one question does not apply in another question.[/i] 1.Suppose that $Area( \vartriangle AJB) =6$ and that $Area(\vartriangle BJC) = 9$. Determine $Area(\vartriangle APD)$. 2. Find all points $Q$ on the plane of the trapezium such that $Area(\vartriangle AQB) = Area(\vartriangle DQC)$. 3. Prove that $PJ$ is the angle bisector of $\angle APD$.

A triangle with angles of $30, 70$ and $80$ degrees is given. Cut it by a straight line into two triangles in such a way that an angle bisector in one of these triangles and a median in the other one drawn from two endpoints of the cutting segment are parallel to each other. (It suffices to find one such cutting.) (A. Shapovalov )

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.

In trapezoid $ABCD$, $AD$ is parallel to $BC$. Knowing that $AB=AD+BC$, prove that the bisector of $\angle A$ also bisects $CD$.

In a rectangle $ABCD, M$ and $N$ are the midpoints of $AD$ and $BC$ respectively and $P$ is a point on line $CD$. The line $PM$ meets $AC$ at $Q$. Prove that MN bisects the angle $\angle QNP$.

Let $\omega,\omega_1,\omega_2$ be three mutually tangent circles such that $\omega_1,\omega_2$ are externally tangent at $P$, $\omega_1,\omega$ are internally tangent at $A$, and $\omega,\omega_2$ are internally tangent at $B$. Let $O,O_1,O_2$ be the centers of $\omega,\omega_1,\omega_2$, respectively. Given that $X$ is the foot of the perpendicular from $P$ to $AB$, prove that $\angle{O_1XP}=\angle{O_2XP}$. [i]David Yang.[/i]

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 $BF + DE = AE$.

Let $ABC$ be an acute triangle with $AB < AC$. The points $A_1$ and $A_2$ are located on the line $BC$ so that $AA_1$ and $AA_2$ are the inner and outer angle bisectors at $A$ for the triangle $ABC$. Let $A_3$ be the mirror image $A_2$ with respect to $C$, and let $Q$ be a point on $AA_1$ such that $\angle A_1QA_3 = 90^o$. Show that $QC // AB$.

In the $ ABC$ triangle, the bisector of $A $ intersects the $ [BC] $ at the point $ A_ {1} $ , and the circle circumscribed to the triangle $ ABC $ at the point $ A_ {2} $. Similarly are defined $ B_ {1} $ and $ B_ {2} $ , as well as $ C_ {1} $ and $ C_ {2} $. Prove that $$ \frac {A_{1}A_{2}}{BA_{2} + A_{2}C} + \frac {B_{1}B_{2}}{CB_{2} + B_{2}A} + \frac {C_{1}C_{2}}{AC_{2} + C_{2}B} \geq \frac {3}{4}$$

In triangle $\triangle ABC$ , $I$ is the incenter and $M$ is the midpoint of arc $(BC)$ in the circumcircle of $(ABC)$not containing $A$. Let $X$ be an arbitrary point on the external angle bisector of $A$. Let $BX \cap (BIC) = T$. $Y$ lies on $(AXC)$ , different from $A$ , st $MA=MY$ . Prove that $TC || AY$ (Assume that $X$ is not on $(ABC)$ or $BC$)