Two circles $(O)$ and $(O')$ intersect at $A$ and $B$. Take two points $P,Q$ on $(O)$ and $(O')$, respectively, such that $AP=AQ$. The line $PQ$ intersects $(O)$ and $(O')$ respectively at $M,N$. Let $E,F$ respectively be the centers of the two arcs $BP$ and $BQ$ (which don't contains $A$). Prove that $MNEF$ is a cyclic quadrilateral.

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.

Let $ABCD$ be a convex quadrilateral with $m(\widehat{ADC}) = 90^{\circ}$. The line through $D$ which is parallel to $BC$ meets $AB$ at $E$. If $m(\widehat{DAC}) = m(\widehat{DAE})$, $|AB|=3$ and $|AC|=4$, then $|AE| = ?$ $\textbf{(A)}\ \frac56 \qquad\textbf{(B)}\ \frac13 \qquad\textbf{(C)}\ \frac12 \qquad\textbf{(D)}\ 1 \qquad\textbf{(E)}\ \frac34$

$ABCD$ is a trapezoid with $AB$ parallel to $CD$. The external bisectors of the angles at $ B$ and $C$ intersect at $ P$. The external bisectors of the angles at $ A$ and $D$ intersect at $Q$. Show that the length of $PQ$ is equal to half the perimeter of the trapezoid $ABCD$.

Let $\triangle ABC$ be an acute angled triangle and let $k$ be its circumscribed circle. A point $O$ is given in the interior of the triangle, such that $CE=CF$, where $E$ and $F$ are on $k$ and $E$ lies on $AO$ while $F$ lies on $BO$. Prove that $O$ is on the angle bisector of $\angle ACB$ if and only if $AC=BC$.

Let $ ABC$ be an acute triangle and $ CL$ be the angle bisector of $ \angle ACB$. The point $ P$ lies on the segment $CL$ such that $ \angle APB\equal{}\pi\minus{}\frac{_1}{^2}\angle ACB$. Let $ k_1$ and $ k_2$ be the circumcircles of the triangles $ APC$ and $ BPC$. $ BP\cap k_1\equal{}Q, AP\cap k_2\equal{}R$. The tangents to $ k_1$ at $ Q$ and $ k_2$ at $ B$ intersect at $ S$ and the tangents to $ k_1$ at $ A$ and $ k_2$ at $ R$ intersect at $ T$. Prove that $ AS\equal{}BT.$

Let ABCD be a parallelogram, M the midpoint of AB and N the intersection of CD and the angle bisector of ABC. Prove that CM and BN are perpendicular iff AN is the angle bisector of DAB.

Let $ABCDEF$ be a regular hexagon and $M\in (DE)$, $N\in(CD)$ such that $m (\widehat {AMN}) = 90^\circ$ and $AN = CM \sqrt {2}$. Find the value of $\frac{DM}{ME}$.

Given a triangle $ABC$, the internal and external bisectors of angle $A$ intersect $BC$ at points $D$ and $E$ respectively. Let $F$ be the point (different from $A$) where line $AC$ intersects the circle $w$ with diameter $DE$. Finally, draw the tangent at $A$ to the circumcircle of triangle $ABF$, and let it hit $w$ at $A$ and $G$. Prove that $AF=AG$.

Let $ABC$ be a triangle and let $O$ be its circumcentre. The internal and external bisectrices of the angle $BAC$ meet the line $BC$ at points $D$ and $E$, respectively. Let further $M$ and $L$ respectively denote the midpoints of the segments $BC$ and $DE$. The circles $ABC$ and $ALO$ meet again at point $N$. Show that the angles $BAN$ and $CAM$ are equal.

Let $ABC$ be a triangle where $M$ is the midpoint of $\overline{AC}$, and $\overline{CN}$ is the angle bisector of $\angle ACB$ with $N$ on $\overline{AB}$. Let $X$ be the intersection of the median $\overline{BM}$ and the bisector $\overline{CN}$. In addition $\bigtriangleup BXN$ is equilateral and $AC=2$. What is $BN^2$? $\textbf{(A)}\ \frac{10-6\sqrt{2}}{7} \qquad\textbf{(B)}\ \frac{2}{9} \qquad\textbf{(C)}\ \frac{5\sqrt{2} - 3\sqrt{3}}{8} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{6} \qquad\textbf{(E)}\ \frac{3\sqrt{3} - 4}{5}$.

Points $H$ and $L$ are, respectively, the feet of the altitude and the angle bisector drawn from the vertex $A$ of the triangle $ABC$, $K$ is the touchpoint of the circle inscribed in the triangle $ABC$ with the side $BC$. Under what conditions will $AK$ be the bisector of the angle $\angle LAH$? (Hryhorii Filippovskyi)

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 $ABC$ be an acute-angled triangle. The line through $C$ perpendicular to $AC$ meets the external angle bisector of $\angle ABC$ at $D$. Let $H$ be the foot of the perpendicular from $D$ onto $BC$. The point $K$ is chosen on $AB$ so that $KH \parallel AC$. Let $M$ be the midpoint of $AK$. Prove that $MC = MB + BH$. Giorgi Arabidze, Georgia,

If $ AD$ is the altitude, $ BE$ the angle bisector, and $ CF$ the median of a triangle $ ABC$, prove that $ AD,BE,$ and $ CF$ are concurrent if and only if: $ a^2(a\minus{}c)\equal{}(b^2\minus{}c^2)(a\plus{}c),$ where $ a,b,c$ are the lengths of the sides $ BC,CA,AB$, respectively.

Given triangle $ABC$, with $AC> BC$, and the it's circumcircle centered at $O$. Let $M$ be the point on the circumcircle of triangle $ABC$ so that $CM$ is the bisector of $\angle ACB$. Let $\Gamma$ be a circle with diameter $CM$. The bisector of $BOC$ and bisector of $AOC$ intersect $\Gamma$ at $P$ and $Q$, respectively. If $K$ is the midpoint of $CM$, prove that $P, Q, O, K$ lie at one point of the circle.

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 \[FB \cdot FA+CP^2 = CF^2 \iff AD \cdot BE = PD^2.\]

Let $ABC$ be a triangle such that $\angle{ABC} = 2\angle{BCA}$ and $\angle{CAB}>90^\circ$. Let $M$ be the midpoint of $BC$. The line perpendicular to $AC$ that passes through $C$ cuts the line $AB$ at point $D$. Show that $\angle{AMB} = \angle{DMC}$.

In a triangle $ABC$ , $D$ is a point on $BC$ such that $AD$ is the internal bisector of $\angle A$ . Now Suppose $\angle B$=$2\angle C$ and $CD=AB$ . Prove that $\angle A=72^0$.

Let $ABC$ be an isosceles triangle with $AB=AC$. Consider a variable point $P$ on the extension of the segment $BC$ beyound $B$ (in other words, $P$ lies on the line $BC$ such that the point $B$ lies inside the segment $PC$). Let $r_{1}$ be the radius of the incircle of the triangle $APB$, and let $r_{2}$ be the radius of the $P$-excircle of the triangle $APC$. Prove that the sum $r_{1}+r_{2}$ of these two radii remains constant when the point $P$ varies. [i]Remark.[/i] The $P$-excircle of the triangle $APC$ is defined as the circle which touches the side $AC$ and the [i]extensions[/i] of the sides $AP$ and $CP$.

$ABC$ is a triangle. Points $D, E, F$ are chosen on $BC, CA, AB$ such that $B$ is equidistant from $D$ and $F$, and $C$ is equidistant from $D$ and $E$. Show that the circumcenter of $AEF$ lies on the bisector of $EDF$.

Two circles $ \Omega_{1}$ and $ \Omega_{2}$ are externally tangent to each other at a point $ I$, and both of these circles are tangent to a third circle $ \Omega$ which encloses the two circles $ \Omega_{1}$ and $ \Omega_{2}$. The common tangent to the two circles $ \Omega_{1}$ and $ \Omega_{2}$ at the point $ I$ meets the circle $ \Omega$ at a point $ A$. One common tangent to the circles $ \Omega_{1}$ and $ \Omega_{2}$ which doesn't pass through $ I$ meets the circle $ \Omega$ at the points $ B$ and $ C$ such that the points $ A$ and $ I$ lie on the same side of the line $ BC$. Prove that the point $ I$ is the incenter of triangle $ ABC$. [i]Alternative formulation.[/i] Two circles touch externally at a point $ I$. The two circles lie inside a large circle and both touch it. The chord $ BC$ of the large circle touches both smaller circles (not at $ I$). The common tangent to the two smaller circles at the point $ I$ meets the large circle at a point $ A$, where the points $ A$ and $ I$ are on the same side of the chord $ BC$. Show that the point $ I$ is the incenter of triangle $ ABC$.

$D$ and $E$ are points on the sides $BC$ and $AC$ of a triangle $ABC$ such that $AD$ and $BE$ are angle bisectors of the triangle $ABC$. If $DE$ bisects $\angle ADC$, find $\angle A$. (SI Tokarev)

In triangle $ ABC$, $ AC \equal{} 13, BC \equal{} 14,$ and $ AB\equal{}15$. Points $ M$ and $ D$ lie on $ AC$ with $ AM\equal{}MC$ and $ \angle ABD \equal{} \angle DBC$. Points $ N$ and $ E$ lie on $ AB$ with $ AN\equal{}NB$ and $ \angle ACE \equal{} \angle ECB$. Let $ P$ be the point, other than $ A$, of intersection of the circumcircles of $ \triangle AMN$ and $ \triangle ADE$. Ray $ AP$ meets $ BC$ at $ Q$. The ratio $ \frac{BQ}{CQ}$ can be written in the form $ \frac{m}{n}$, where $ m$ and $ n$ are relatively prime positive integers. Find $ m\minus{}n$.

the edges of triangle $ABC$ are $a,b,c$ respectively,$b<c$,$AD$ is the bisector of $\angle A$,point $D$ is on segment $BC$. (1)find the property $\angle A$,$\angle B$,$\angle C$ have,so that there exists point $E,F$ on $AB,AC$ satisfy $BE=CF$,and $\angle NDE=\angle CDF$ (2)when such $E,F$ exist,express $BE$ with $a,b,c$