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.

The triangle $ABC$ is drawn on the board such that $AB + AC = 2BC$. The bisectors $AL_1, BL_2, CL_3$ were drawn in this triangle, after which everything except the points $L_1, L_2, L_3$ was erased. Use a compass and a ruler to reconstruct triangle $ABC$.

Let $\triangle ABC$ with $\angle BAC=120 ^\circ$. Let $D, E, F$ points on sides $BC, CA, AB$, respectively, such that $AD, BE, CF$ are angle bisectors on $\triangle ABC$. \\ Prove that $\triangle ABC$ is isosceles if and only if $\triangle DEF$ is right-angled isosceles.

Triangle $ABC$ is right angled such that $\angle ACB=90^{\circ}$ and $\frac {AC}{BC} = 2$. Let the line parallel to side $AC$ intersects line segments $AB$ and $BC$ in $M$ and $N$ such that $\frac {CN}{BN} = 2$. Let $O$ be the intersection point of lines $CM$ and $AN$. On segment $ON$ lies point $K$ such that $OM+OK=KN$. Let $T$ be the intersection point of angle bisector of $\angle ABC$ and line from $K$ perpendicular to $AN$. Determine value of $\angle MTB$.

In convex quadrilateral $ ABCD$, $ AB \equal{} BC$ and $ AD \equal{} DC$. Point $ E$ lies on segment $ AB$ and point $ F$ lies on segment $ AD$ such that $ B$, $ E$, $ F$, $ D$ lie on a circle. Point $ P$ is such that triangles $ DPE$ and $ ADC$ are similar and the corresponding vertices are in the same orientation (clockwise or counterclockwise). Point $ Q$ is such that triangles $ BQF$ and $ ABC$ are similar and the corresponding vertices are in the same orientation. Prove that points $ A$, $ P$, $ Q$ are collinear.

The inscribed circle of an acute $\Delta ABC$ is tangent to $AB$ and $AC$ in $K$ and $L$ respectively. The altitude $AH$ intersects the angle bisectors of $\angle ABC$ and $\angle ACB$ in $P$ and $Q$ respectively. Prove that the middle point $M$ of $AH$ lies on the radical axis of the circumscribed circles of $\Delta KPB$ and $\Delta LQC$.

Let $ABCD$ be a convex cyclic quadrilateral. Prove that: $a)$ the number of points on the circumcircle of $ABCD$, like $M$, such that $\frac{MA}{MB}=\frac{MD}{MC}$ is $4$. $b)$ The diagonals of the quadrilateral which is made with these points are perpendicular to each other.

Let $ABC$ be a right triangle with $m(\widehat{A})=90^\circ$. Let $APQR$ be a square with area $9$ such that $P\in [AC]$, $Q\in [BC]$, $R\in [AB]$. Let $KLMN$ be a square with area $8$ such that $N,K\in [BC]$, $M\in [AB]$, and $L\in [AC]$. What is $|AB|+|AC|$? $ \textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 12 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16 $

Bisectors $AI$ and $CI$ meet the circumcircle of triangle $ABC$ at points $A_1, C_1$ respectively. The circumcircle of triangle $AIC_1$ meets $AB$ at point $C_0$; point $A_0$ is defined similarly. Prove that $A_0, A_1, C_0, C_1$ are collinear.

Let $A$ and $B$ be points on the circle $k$ with center $O$, so that $AB> AO$. Let $C$ be the intersection of the bisectors of $\angle OAB$ and $k$, different from $A$. Let $D$ be the intersection of the straight line $AB$ with the circumcircle of the triangle $OBC$, different from $B$. Show that $AD = AO$ .

Triangle $ABC$ is inscribed in circle $\omega$. A circle with chord $BC$ intersects segments $AB$ and $AC$ again at $S$ and $R$, respectively. Segments $BR$ and $CS$ meet at $L$, and rays $LR$ and $LS$ intersect $\omega$ at $D$ and $E$, respectively. The internal angle bisector of $\angle BDE$ meets line $ER$ at $K$. Prove that if $BE = BR$, then $\angle ELK = \tfrac{1}{2} \angle BCD$. [i]Proposed by Evan Chen[/i]

Let $ABC$ be a triangle inscribed in the circle $(O)$. Let $I$ be the center of the circle inscribed in the triangle and $D$ the point of contact of the circle inscribed with the side $BC$. Let $M$ be the second intersection point of the bisector $AI$ with the circle $(O)$ and let $P$ be the point where the line $DM$ intersects the circle $(O)$ . Show that $PA \perp PI$.

Given a triangle $ABC$ inscribed by circumcircle $(O)$. The angles at $B,C$ are acute angle. Let $M$ on the arc $BC$ that doesn't contain $A$ such that $AM$ is not perpendicular to $BC$. $AM$ meets the perpendicular bisector of $BC$ at $T$. The circumcircle $(AOT)$ meets $(O)$ at $N$ ($N\ne A$). a) Prove that $\angle{BAM}=\angle{CAN}$. b) Let $I$ be the incenter and $G$ be the foor of the angle bisector of $\angle{BAC}$. $AI,MI,NI$ intersect $(O)$ at $D,E,F$ respectively. Let ${P}=DF\cap AM, {Q}=DE\cap AN$. The circle passes through $P$ and touches $AD$ at $I$ meets $DF$ at $H$ ($H\ne D$).The circle passes through $Q$ and touches $AD$ at $I$ meets $DE$ at $K$ ($K\ne D$). Prove that the circumcircle $(GHK)$ touches $BC$.

Let $A_1A_2A_3\ldots A_n$ be a regular $n$-gon. Let $B_1$ and $B_{n-1}$ be the midpoints of its sides $A_1A_2$ and $A_{n-1}A_n$. Also, for every $i\in\left\{2,3,4,\ldots ,n-2\right\}$. Let $S$ be the point of intersection of the lines $A_1A_{i+1}$ and $A_nA_i$, and let $B_i$ be the point of intersection of the angle bisector bisector of the angle $\measuredangle A_iSA_{i+1}$ with the segment $A_iA_{i+1}$. Prove that $\sum_{i=1}^{n-1} \measuredangle A_1B_iA_n=180^{\circ}$. [i]Proposed by Dusan Dukic, Serbia and Montenegro[/i]

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 triangle $ ABC$ with $ \left|BC\right|>\left|BA\right|$, $ D$ is a point inside the triangle such that $ \angle ABD\equal{}\angle DBC$, $ \angle BDC\equal{}150{}^\circ$ and $ \angle DAC\equal{}60{}^\circ$. What is the measure of $ \angle BAD$? $\textbf{(A)}\ 45 \qquad\textbf{(B)}\ 50 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 75 \qquad\textbf{(E)}\ 80$