Let $ABCD$ be a convex quadrilateral with $\angle{DAC}= \angle{BDC}= 36^\circ$ , $\angle{CBD}= 18^\circ$ and $\angle{BAC}= 72^\circ$. The diagonals and intersect at point $P$ . Determine the measure of $\angle{APD}$.

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.

It is known that $ABCD$ is a parallelogram. The point $E$ is taken so that $BCED$ is a cyclic quadrilateral. Let $\ell$ be a line that passes through $A$, intersects the segment $DC$ at point $F$ and intersects the extension of the line $BC$ at $G$. Given $EF = EG = EC$. Prove that $\ell$ is the bisector of the angle $\angle BAD$.

Two circles are externally tangent to each other at a point $A$ and internally tangent to a third circle $\Gamma$ at points $B$ and $C.$ Let $D$ be the midpoint of the secant of $\Gamma$ which is tangent to the smaller circles at $A.$ Show that $A$ is the incenter of the triangle $BCD$ if the centers of the circles are not collinear.

In an acute triangle $ABC$, the segment $CD$ is an altitude and $H$ is the orthocentre. Given that the circumcentre of the triangle lies on the line containing the bisector of the angle $DHB$, determine all possible values of $\angle CAB$.

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 $BC$.

Let a cyclic quadrilateral $ ABCD$, $ AC \cap BD \equal{} E$ and let a circle $ \Gamma$ internally tangent to the arch $ BC$ (that not contain $ D$) in $ T$ and tangent to $ BE$ and $ CE$. Call $ R$ the point where the angle bisector of $ \angle ABC$ meet the angle bisector of $ \angle BCD$ and $ S$ the incenter of $ BCE$. Prove that $ R$, $ S$ and $ T$ are collinear. [i](Gabriel Giorgieri)[/i]

Does a parallelogram exist such that all pairwise meets of bisectors of its angles are situated outside it?

Let $ABC$ be a triangle with area $1$ and $P$ the middle of the side $[BC]$. $M$ and $N$ are two points of $[AB]-\left \{ A,B \right \} $ and $[AC]-\left \{ A,C \right \}$ respectively such that $AM=2MB$ and$CN=2AN$. The two lines $(AP)$ and $(MN)$ intersect in a point $D$. Find the area of the triangle $ADN$.

Let $ABC$ be an equilateral triangle whose angle bisectors of $B$ and $C$ intersect at $D$. Perpendicular bisectors of $BD$ and $CD$ intersect $BC$ at points $E$ and $Z$ respectively. a) Prove that $BE=EZ=ZC$. b) Find the ratio of the areas of the triangles $BDE$ to $ABC$

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]

Points $X$ and $Y$ are the midpoints of arcs $AB$ and $BC$ of the circumscribed circle of triangle $ABC$. Point $T$ lies on side $AC$. It turned out that the bisectors of the angles $ATB$ and $BTC$ pass through points $X$ and $Y$ respectively. What angle $B$ can be in triangle $ABC$?

In a triangle $ABC$, let $D$ and $E$ be the midpoints of $AB$ and $AC$, respectively, and let $F$ be the foot of the altitude through $A$. Show that the line $DE$, the angle bisector of $\angle ACB$ and the circumcircle of $ACF$ pass through a common point. [b]Alternate version:[/b] In a triangle $ABC$, let $D$ and $E$ be the midpoints of $AB$ and $AC$, respectively. The line $DE$ and the angle bisector of $\angle ACB$ meet at $G$. Show that $\angle AGC$ is a right angle.

We are given a right-angled triangle $MNP$ with right angle in $P$. Let $k_M$ be the circle with center $M$ and radius $MP$, and let $k_N$ be the circle with center $N$ and radius $NP$. Let $A$ and $B$ be the common points of $k_M$ and the line $MN$, and let $C$ and $D$ be the common points of $k_N$ and the line $MN$ with with $C$ between $A$ and $B$. Prove that the line $PC$ bisects the angle $\angle APB$.

A triangle $ABC$ is given in which $\angle BAC = 40^o$. and $\angle ABC = 20^o$. Find the length of the angle bisector drawn from the vertex $C$, if it is known that the sides $AB$ and $BC$ differ by $4$ centimeters.