Square $ BDEC$ with center $ F$ is constructed to the out of triangle $ ABC$ such that $ \angle A \equal{} 90{}^\circ$, $ \left|AB\right| \equal{} \sqrt {12}$, $ \left|AC\right| \equal{} 2$. If $ \left[AF\right]\bigcap \left[BC\right] \equal{} \left\{G\right\}$ , then $ \left|BG\right|$ will be $\textbf{(A)}\ 6 \minus{} 2\sqrt {3} \qquad\textbf{(B)}\ 2\sqrt {3} \minus{} 1 \qquad\textbf{(C)}\ 2 \plus{} \sqrt {3} \\ \qquad\textbf{(D)}\ 4 \minus{} \sqrt {3} \qquad\textbf{(E)}\ 5 \minus{} 2\sqrt {2}$

Given an equilateral $\triangle ABC$, find the locus of points $P$ such that $\angle APB=\angle BPC$.

On an acute triangle $ABC$ we draw the internal bisector of $<ABC$, $BE$, and the altitude $AD$, ($D$ on $BC$), show that $<CDE$ it's bigger than 45 degrees.

Given is $\triangle ABC$ such that $\angle A = 57^o, \angle B = 61^o$ and $\angle C = 62^o$. Which segment is longer: the angle bisector through $A$ or the median through $B$? (N.Beluhov)

A line $r$ contains the points $A$, $B$, $C$, $D$ in that order. Let $P$ be a point not in $r$ such that $\angle{APB} = \angle{CPD}$. Prove that the angle bisector of $\angle{APD}$ intersects the line $r$ at a point $G$ such that: $\frac{1}{GA} + \frac{1}{GC} = \frac{1}{GB} + \frac{1}{GD}$

Given a triangle $ \,ABC,\,$ let $ \,I\,$ be the center of its inscribed circle. The internal bisectors of the angles $ \,A,B,C\,$ meet the opposite sides in $ \,A^{\prime },B^{\prime },C^{\prime }\,$ respectively. Prove that \[ \frac {1}{4} < \frac {AI\cdot BI\cdot CI}{AA^{\prime }\cdot BB^{\prime }\cdot CC^{\prime }} \leq \frac {8}{27}. \]

Let $ABC$ be an isosceles triangle with $AC=BC$, let $M$ be the midpoint of its side $AC$, and let $Z$ be the line through $C$ perpendicular to $AB$. The circle through the points $B$, $C$, and $M$ intersects the line $Z$ at the points $C$ and $Q$. Find the radius of the circumcircle of the triangle $ABC$ in terms of $m = CQ$.

An equilateral triangle with side length $6$ has a square of side length $6$ attached to each of its edges as shown. The distance between the two farthest vertices of this figure (marked $A$ and $B$ in the figure) can be written as $m + \sqrt{n}$ where $m$ and $n$ are positive integers. Find $m + n$. [asy] draw((0,0)--(1,0)--(1/2,sqrt(3)/2)--cycle); draw((1,0)--(1+sqrt(3)/2,1/2)--(1/2+sqrt(3)/2,1/2+sqrt(3)/2)--(1/2,sqrt(3)/2)); draw((0,0)--(-sqrt(3)/2,1/2)--(-sqrt(3)/2+1/2,1/2+sqrt(3)/2)--(1/2,sqrt(3)/2)); dot((-sqrt(3)/2+1/2,1/2+sqrt(3)/2)); label("A", (-sqrt(3)/2+1/2,1/2+sqrt(3)/2), N); draw((1,0)--(1,-1)--(0,-1)--(0,0)); dot((1,-1)); label("B", (1,-1), SE); [/asy]

a) Let $UVW$ , $U'V'W'$ be two triangles such that $ VW = V'W' , UV = U'V' , \angle WUV = \angle W'U'V'.$ Prove that the angles $\angle VWU , \angle V'W'U'$ are equal or supplementary. b) $ABC$ is a triangle where $\angle A$ is [b]obtuse[/b]. take a point $P$ inside the triangle , and extend $AP,BP,CP$ to meet the sides $BC,CA,AB$ in $K,L,M$ respectively. Suppose that $PL = PM .$ 1) If $AP$ bisects $\angle A$ , then prove that $AB = AC$ . 2) Find the angles of the triangle $ABC$ if you know that $AK,BL,CM$ are angle bisectors of the triangle $ABC$ and that $2AK = BL$.

Let $\triangle ABC$ be a triangle such that $\angle C$ and $\angle B$ are acute. Let $D$ be a variable point on $BC$ such that $D\neq B, C$ and $AD$ is not perpendicular to $BC.$ Let $d$ be the line passing through $D$ and perpendicular to $BC.$ Assume $d \cap AB= E, d \cap AC =F.$ If $M, N, P$ are the incentres of $\triangle AEF, \triangle BDE,\triangle CDF.$ Prove that $A, M, N, P$ are concyclic if and only if $d$ passes through the incentre of $\triangle ABC.$

In an acute triangle $ABC$ the angle $C$ is greater than the angle $A$. Let $AE$ be a diameter of the circumcircle of the triangle. Let the intersection point of the ray $AC$ and the tangent of the circumcircle through the vertex $B$ be $K$. The perpendicular to $AE$ through $K$ intersects the circumcircle of the triangle $BCK$ for the second time at point $D$. Prove that $CE$ bisects the angle $BCD$.

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 a triangle $ABC$ with a right angle at $C$, the angle bisector $AL$ (where $L$ is on segment $BC$) intersects the altitude $CH$ at point $K$. The bisector of angle $BCH$ intersects segment $AB$ at point $M$. Prove that $CK=ML$

Let $ABC$ be an acute-angled triangle with $\angle ACB = 60^o$ , and its heights $AD$ and $BE$ intersect at point $H$. Prove that the circumcenter of triangle $ABC$ lies on a line bisecting the angles $AHE$ and $BHD$.

In the scalene triangle $ABC$,$\angle ACB=60$ and $\Omega$ is its cirumcirle.On the bisectors of the angles $BAC$ and $CBA$ points $A^\prime$,$B^\prime$ are chosen respectively such that $AB^\prime \parallel BC$ and $BA^\prime \parallel AC$.$A^\prime B^\prime$ intersects with $\Omega$ at $D,E$.Prove that triangle $CDE$ is isosceles.(A. Kuznetsov)

Let a line $\ell$ intersect the line $AB$ at $F$, the sides $AC$ and $BC$ of a triangle $ABC$ at $D$ and $E$, respectively and the internal bisector of the angle $BAC$ at $P$. Suppose that $F$ is at the opposite side of $A$ with respect to the line $BC$, $CD = CE$ and $P$ is in the interior the triangle $ABC$. Prove that \[FB \cdot FA+CP^2 = CF^2 \iff AD \cdot BE = PD^2.\]

In triangle $ABC$, $AD$ and $AH$ are the angle bisector and the altitude of vertex $A$, respectively. The perpendicular bisector of $AD$, intersects the semicircles with diameters $AB$ and $AC$ which are drawn outside triangle $ABC$ in $X$ and $Y$, respectively. Prove that the quadrilateral $XYDH$ is concyclic. [i]Proposed by Mahan Malihi[/i]

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]

A right-angled triangle has an angle equal to $30^\circ.$ Prove that one of the bisectors of the triangle is twice as short as another one. [i]Egor Bakaev[/i]

Let $ABC$ be a triangle with $AC > AB$. Let $P$ be the intersection point of the perpendicular bisector of $BC$ and the internal angle bisector of $\angle{A}$. Construct points $X$ on $AB$ (extended) and $Y$ on $AC$ such that $PX$ is perpendicular to $AB$ and $PY$ is perpendicular to $AC$. Let $Z$ be the intersection point of $XY$ and $BC$. Determine the value of $\frac{BZ}{ZC}$.

In a triangle $ABC$ the bisectors of angles $A$ and $C$ meet the opposite sides at $D$ and $E$ respectively. Show that if the angle at $B$ is greater than $60^\circ$, then $AE +CD <AC$.

The orthocenter of an acute triangle $ABC$ is $H$. Let $K$ and $P$ be the midpoints of lines $BC$ and $AH$, respectively. The angle bisector drawn from the vertex $A$ of the triangle $ABC$ intersects with line $KP$ at $D$. Prove that $HD\perp AD$.

Let $ABC$ be a scalene triangle and $\omega$ is your incircle. The sides $BC,CA$ and $AB$ are tangents to $\omega$ in $X,Y,Z$ respectively. Let $M$ be the midpoint of $BC$ and $D$ is the intersection point of $BC$ with the angle bisector of $\angle BAC$. Prove that $\angle BAX=\angle MAC$ if and only if $YZ$ passes by the midpoint of $AD$.

In $\triangle ABC$, $AB = 40$, $BC = 60$, and $CA = 50$. The angle bisector of $\angle A$ intersects the circumcircle of $\triangle ABC$ at $A$ and $P$. Find $BP$. [i]Proposed by Eugene Chen[/i]

Let $\triangle ABC$ be a triangle and $\omega$ its circumcircle. The exterior angle bisector of $\angle BAC$ intersects $\omega$ at point $D$. Let $X$ be the foot of the altitude from $C$ to $AD$ and let $F$ be the intersection of the internal angle bisector of $\angle BAC$ and $BC$. Show that $BX$ bisects segment $AF$.