Let $ Q$ be the centre of the inscribed circle of a triangle $ ABC.$ Prove that for any point $ P,$ \[ a(PA)^2 \plus{} b(PB)^2 \plus{} c(PC)^2 \equal{} a(QA)^2 \plus{} b(QB)^2 \plus{} c(QC)^2 \plus{} (a \plus{} b \plus{} c)(QP)^2, \] where $ a \equal{} BC, b \equal{} CA$ and $ c \equal{} AB.$

Let $ ABCD$ be the inscribed quadrilateral with the circumcircle $ \omega$.Let $ \zeta$ be another circle that internally tangent to $ \omega$ and to the lines $ BC$ and $ AD$ at points $ M,N$ respectively.Let $ I_1,I_2$ be the incenters of the $ \triangle ABC$ and $ \triangle ABD$.Prove that $ M,I_1,I_2,N$ are collinear.

Circles $\Omega $ and $\omega $ are tangent at a point $P$ ($\omega $ lies inside $\Omega $). A chord $AB$ of $\Omega $ is tangent to $\omega $ at $C;$ the line $PC$ meets $\Omega $ again at $Q.$ Chords $QR$ and $QS$ of $ \Omega $ are tangent to $\omega .$ Let $I,X,$ and $Y$ be the incenters of the triangles $APB,$ $ARB,$ and $ASB,$ respectively. Prove that $\angle PXI+\angle PYI=90^{\circ }.$

Let $ABC$ be an acute triangle with $\angle B > \angle C$. On the circle $\mathcal{C}(O, R)$ circumscribed to this triangle points $D, E, J, K, S$ are chosen such that $A, E, J$ and $K$ are on the same side of the line $BC$, the diameter $DE$ is perpendicular on the chord $BC$, $S\in \overarc{EK},\overarc{AE}=\overarc{BJ}=\overarc{CK}=\dfrac{1}{4}\overarc{CE}$ . Let $\{F\}=AC\cap DE, \{M\}=BK\cap AD, \{P\}=BK\cap AC$ and $\{Q\}=CJ\cap BF$. If $\angle SMK =30^{\circ}$ and $\angle AQP = 90^{\circ}$, show that the line $MS$ is tangent to the circumscribed circle of triangle $AOF$.

Let $K$ be a point inside an acute triangle $ ABC $, such that $ BC $ is a common tangent of the circumcircles of $ AKB $ and $ AKC$. Let $ D $ be the intersection of the lines $ CK $ and $ AB $, and let $ E $ be the intersection of the lines $ BK $ and $ AC $ . Let $ F $ be the intersection of the line $BC$ and the perpendicular bisector of the segment $DE$. The circumcircle of $ABC$ and the circle $k$ with centre $ F$ and radius $FD$ intersect at points $P$ and $Q$. Prove that the segment $PQ$ is a diameter of $k$.

The points $ P, Q$ are given on the plane, which are the points of intersection of the angle bisector $AL$ of some triangle $ABC$ with an inscribed circle, and the point $W$ is the intersection of the angle bisector $AL$ with a circumscribed circle other than the vertex $A$. a) Find the geometric locus of the possible location of the vertex $A$ of the triangle $ABC$. b) Find the geometric locus of the possible location of the vertex $B$ of the triangle $ABC$.

Let $ABC$ be a triangle and $AB<AC<BC$. Let $D,E$ be points on the side $BC$ and the line $AB$, respectively ($A$ is between $B,E$) such that $BD=BE=AC$. The circumcircle of $\Delta BED$ meets the side $AC$ at $P$ and $BP$ meets the circumcircle of $\Delta ABC$ at $Q$. Prove that: \[ AQ+CQ=BP. \]

Let $ ABCD$ be a rhombus with $ \angle BAD \equal{} 60^{\circ}$. Points $ S$ and $ R$ are chosen inside the triangles $ ABD$ and $ DBC$, respectively, such that $ \angle SBR \equal{} \angle RDS \equal{} 60^{\circ}$. Prove that $ SR^2\geq AS\cdot CR$.

Let $ABC$ be a triangle and $AD,BE,CF$ be cevians concurrent at a point $P$. Suppose each of the quadrilaterals $PDCE,PEAF$ and $PFBD$ has both circumcircle and incircle. Prove that $ABC$ is equilateral and $P$ coincides with the center of the triangle.

Let $\omega$ be the circumcircle of right-angled triangle $ABC$ ($\angle A = 90^{\circ}$). The tangent to $\omega$ at point $A$ intersects the line $BC$ at point $P$. Suppose that $M$ is the midpoint of the minor arc $AB$, and $PM$ intersects $\omega$ for the second time in $Q$. The tangent to $\omega$ at point $Q$ intersects $AC$ at $K$. Prove that $\angle PKC = 90^{\circ}$. [i]Proposed by Davood Vakili[/i]

Let a convex quadrilateral $ ABCD$ fixed such that $ AB \equal{} BC$, $ \angle ABC \equal{} 80, \angle CDA \equal{} 50$. Define $ E$ the midpoint of $ AC$; show that $ \angle CDE \equal{} \angle BDA$ [i](Paolo Leonetti)[/i]

Let $ABC$ be a triangle, and let $D$, $A$, $B$, $E$ be points on line $AB$, in that order, such that $AC=AD$ and $BE=BC$. Let $\omega_1, \omega_2$ be the circumcircles of $\triangle ABC$ and $\triangle CDE$, respectively, which meet at a point $F \neq C$. If the tangent to $\omega_2$ at $F$ cuts $\omega_1$ again at $G$, and the foot of the altitude from $G$ to $FC$ is $H$, prove that $\angle AGH=\angle BGH$. [i]Proposed by David Stoner[/i]

A diagonal of a convex hexagon is called [i]long[/i] if it decomposes the hexagon into two quadrangles. Each pair of [i]long[/i] diagonals decomposes the hexagon into two triangles and two quadrangles. Given is a hexagon with the property, that for each decomposition by two [i]long[/i] diagonals the resulting triangles are both isosceles with the side of the hexagon as base. Show that the hexagon has a circumcircle.

Let $ABC$ be an acute triangle, with $AB \ne AC$. Let $D$ be the midpoint of the line segment $BC$, and let $E$ and $F$ be the projections of $D$ onto the sides $AB$ and $AC$, respectively. If $M$ is the midpoint of the line segment $EF$, and $O$ is the circumcenter of triangle $ABC$, prove that the lines $DM$ and $AO$ are parallel. [hide=PS] As source was given [url=https://artofproblemsolving.com/community/c629086_caucasus_mathematical_olympiad]Caucasus MO[/url], but I was unable to find this problem in the contest collections [/hide]

Let $I$ be the center of a circle inscribed in triangle $ABC$. The circle circumscribed about the triangle $BIC$ intersects lines $AB$ and $AC$ at points $E$ and $F$, respectively. Prove that the line $EF$ touches the circle inscribed in the triangle $ABC$.

Let $ABC$ be an acute-angled triangle and $M$ be the midpoint of the side $BC$. Let $N$ be a point in the interior of the triangle $ABC$ such that $\angle NBA=\angle BAM$ and $\angle NCA=\angle CAM$. Prove that $\angle NAB=\angle MAC$. [i]Gabriel Nagy[/i]

The sides of a triangle are $25,39,$ and $40$. The diameter of the circumscribed circle is: $ \textbf{(A)}\ \frac{133}{3}\qquad\textbf{(B)}\ \frac{125}{3}\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 41\qquad\textbf{(E)}\ 40 $

In the Triangle ABC AB>AC, the extension of the altitude AD with D lying inside BC intersects the circum-circle of the Triangle ABC at P. The circle through P and tangent to BC at D intersects the circum-circle of Triangle ABC at Q distinct from P with PQ=DQ. Prove that AD=BD-DC

In a scalene triangle $\Delta ABC$, $AB<AC$, $PB$ and $PC$ are tangents of the circumcircle $(O)$ of $\Delta ABC$. A point $R$ lies on the arc $\widehat{AC}$(not containing $B$), $PR$ intersects $(O)$ again at $Q$. Suppose $I$ is the incenter of $\Delta ABC$, $ID \perp BC$ at $D$, $QD$ intersects $(O)$ again at $G$. A line passing through $I$ and perpendicular to $AI$ intersects $AB,AC$ at $M,N$, respectively. Prove that, if $AR \parallel BC$, then $A,G,M,N$ are concyclic.

Let $ABC$ be a triangle and let $M$ be the midpoint of the segment $BC$. Let $X$ be a point on the ray $AB$ such that $2 \angle CXA=\angle CMA$. Let $Y$ be a point on the ray $AC$ such that $2 \angle AYB=\angle AMB$. The line $BC$ intersects the circumcircle of the triangle $AXY$ at $P$ and $Q$, such that the points $P, B, C$, and $Q$ lie in this order on the line $BC$. Prove that $PB=QC$. [i]Proposed by Dominik Burek, Poland[/i]

Let $ABC$ be an acute triangle, with $\angle A > 60^\circ$, and let $H$ be it's orthocenter. Let $M$ and $N$ be points on $AB$ and $AC$, respectively, such that $\angle HMB = \angle HNC = 60^\circ$. Also, let $O$ be the circuncenter of $HMN$ and $D$ be a point on the semiplane determined by $BC$ that contains $A$ in such a way that $DBC$ is equilateral. Prove that $H$, $O$ and $D$ are collinear.

In triangle $ABC$ $AA_0$ and $BB_0$ are medians, $AA_1$ and $BB_1$ are altitudes. The circumcircles of triangles $CA_0B_0$ and $CA_1B_1$ meet again in point $M_c$. Points $M_a, M_b$ are defined similarly. Prove that points $M_a, M_b, M_c$ are collinear and lines $AM_a, BM_b, CM_c$ are parallel.

Let $ABC$ be an acute scalene triangle inscribed in a circle \( \Gamma \). The angle bisector of \( \angle BAC \) intersects \( BC \) at \( L \) and \( \Gamma \) at \( S \). The point \( M \) is the midpoint of \( AL \). Let \( AD \) be the altitude in \( \triangle ABC \), and the circumcircle of \( \triangle DSL \) intersects \( \Gamma \) again at \( P \). Let \( N \) be the midpoint of \( BC \), and let \( K \) be the reflection of \( D \) with respect to \( N \). Prove that the triangles \( \triangle MPS \) and \( \triangle ADK \) are similar.

Two circles $\Gamma_1$ and $\Gamma_2$ intersect at points $A$ and $Z$ (with $A\neq Z$). Let $B$ be the centre of $\Gamma_1$ and let $C$ be the centre of $\Gamma_2$. The exterior angle bisector of $\angle{BAC}$ intersects $\Gamma_1$ again at $X$ and $\Gamma_2$ again at $Y$. Prove that the interior angle bisector of $\angle{BZC}$ passes through the circumcenter of $\triangle{XYZ}$. [i]For points $P,Q,R$ that lie on a line $\ell$ in that order, and a point $S$ not on $\ell$, the interior angle bisector of $\angle{PQS}$ is the line that divides $\angle{PQS}$ into two equal angles, while the exterior angle bisector of $\angle{PQS}$ is the line that divides $\angle{RQS}$ into two equal angles.[/i]

Consider parallelogram $ABCD$, with acute angle at $A$, $AC$ and $BD$ intersect at $E$. Circumscribed circle of triangle $ACD$ intersects $AB$, $BC$ and $BD$ at $K$, $L$ and $P$ (in that order). Then, circumscribed circle of triangle $CEL$ intersects $BD$ at $M$. Prove that: $$KD*KM=KL*PC$$