Let $ABCD$ a convex quadrilateral with $AB=BC=CD$, with $AC$ not equal to $BD$ and $E$ be the intersection point of it's diagonals. Prove that $AE=DE$ if and only if $\angle BAD+\angle ADC = 120$.

Let $M$ be an arbitrary point on side $BC$ of triangle $ABC$. $W$ is a circle which is tangent to $AB$ and $BM$ at $T$ and $K$ and is tangent to circumcircle of $AMC$ at $P$. Prove that if $TK||AM$, circumcircles of $APT$ and $KPC$ are tangent together.

Let $ABC$ be an acute-angled triangle with circumcircle $\omega$. Prove that there exists a point $J$ such that for any point $X$ inside $ABC$ if $AX,BX,CX$ intersect $\omega$ in $A_1,B_1,C_1$ and $A_2,B_2,C_2$ be reflections of $A_1,B_1,C_1$ in midpoints of $BC,AC,AB$ respectively then $A_2,B_2,C_2,J$ lie on a circle.

$ABC$ is an acute triangle. Points $M$ and $K$ on side $AC$ are such that $\angle ABM = \angle CBK$. Prove that the circumcenters of triangles $ABM$, $ABK$, $CBM$ and $CBK$ are concyclic. (Author: T. Emelyanova)

Given a triangle $ABC$ where $AC > BC$, $D$ is located on the circumcircle of $ABC$ such that $D$ is the midpoint of the arc $AB$ that contains $C$. $E$ is a point on $AC$ such that $DE$ is perpendicular to $AC$. Prove that $AE = EC + CB$.

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 $BC$.

In the acute-angle triangle $ABC$ we have $\angle ACB = 45^\circ$. The points $A_1$ and $B_1$ are the feet of the altitudes from $A$ and $B$, and $H$ is the orthocenter of the triangle. We consider the points $D$ and $E$ on the segments $AA_1$ and $BC$ such that $A_1D = A_1E = A_1B_1$. Prove that a) $A_1B_1 = \sqrt{ \frac{A_1B^2+A_1C^2}{2} }$; b) $CH=DE$.

Let $ABC$ be a triangle with circumcircle $k$. The tangents at $A$ and $C$ intersect at $T$. The circumcircle of triangle $ABT$ intersects the line $CT$ at $X$ and $Y$ is the midpoint of $CX$. Prove that the lines $AX$ and $BY$ intersect on $k$.

Let $ABC$ be a triangle with $\angle B \le \angle C$, $I$ its incenter and $D$ the intersection point of line $AI$ with side $BC$. Let $M$ and $N$ be points on sides $BA$ and $CA$, respectively, such that $BM = BD$ and $CN = CD$. The circumcircle of triangle $CMN$ intersects again line $BC$ at $P$. Prove that quadrilateral $DIMP$ is cyclic.

Let $ ABCD$ be a trapezoid with parallel sides $ AB > CD$. Points $ K$ and $ L$ lie on the line segments $ AB$ and $ CD$, respectively, so that $AK/KB=DL/LC$. Suppose that there are points $ P$ and $ Q$ on the line segment $ KL$ satisfying \[\angle{APB} \equal{} \angle{BCD}\qquad\text{and}\qquad \angle{CQD} \equal{} \angle{ABC}.\] Prove that the points $ P$, $ Q$, $ B$ and $ C$ are concyclic. [i]Proposed by Vyacheslev Yasinskiy, Ukraine[/i]

Let be given a triangle $ ABC$ with $ BC \equal{} a$, $ CA \equal{} b$, $ AB \equal{} c$. Six distinct points $ A_1$, $ A_2$, $ B_1$, $ B_2$, $ C_1$, $ C_2$ not coinciding with $ A$, $ B$, $ C$ are chosen so that $ A_1$, $ A_2$ lie on line $ BC$; $ B_1$, $ B_2$ lie on $ CA$ and $ C_1$, $ C_2$ lie on $ AB$. Let $ \alpha$, $ \beta$, $ \gamma$ three real numbers satisfy $ \overrightarrow{A_1A_2} \equal{} \frac {\alpha}{a}\overrightarrow{BC}$, $ \overrightarrow{B_1B_2} \equal{} \frac {\beta}{b}\overrightarrow{CA}$, $ \overrightarrow{C_1C_2} \equal{} \frac {\gamma}{c}\overrightarrow{AB}$. Let $ d_A$, $ d_B$, $ d_C$ be respectively the radical axes of the circumcircles of the pairs of triangles $ AB_1C_1$ and $ AB_2C_2$; $ BC_1A_1$ and $ BC_2A_2$; $ CA_1B_1$ and $ CA_2B_2$. Prove that $ d_A$, $ d_B$ and $ d_C$ are concurrent if and only if $ \alpha a \plus{} \beta b \plus{} \gamma c \neq 0$.

Inside the inscribed quadrilateral $ ABCD $, a point $ P $ is marked such that $ \angle PBC = \angle PDA $, $ \angle PCB = \angle PAD $. Prove that there exists a circle that touches the straight lines $ AB $ and $ CD $, as well as the circles circumscribed by the triangles $ ABP $ and $ CDP $.

We are given an acute-angled triangle $ABC$ and a random point $X$ in its interior, different from the centre of the circumcircle $k$ of the triangle. The lines $AX,BX$ and $CX$ intersect $k$ for a second time in the points $A_1,B_1$ and $C_1$ respectively. Let $A_2,B_2$ and $C_2$ be the points that are symmetric of $A_1,B_1$ and $C_1$ in respect to $BC,AC$ and $AB$ respectively. Prove that the circumcircle of the triangle $A_2,B_2$ and $C_2$ passes through a constant point that does not depend on the choice of $X$.

Let $O$ be the center of the circumcircle $\omega$ of an acute-angle triangle $ABC$. A circle $\omega_1$ with center $K$ passes through $A$, $O$, $C$ and intersects $AB$ at $M$ and $BC$ at $N$. Point $L$ is symmetric to $K$ with respect to line $NM$. Prove that $BL \perp AC$.

Given cyclic quadrilateral $ABCD$. Four circles each touching its diagonals and the circumcircle internally are equal. Is $ABCD$ a square? (C.Pohoata, A.Zaslavsky)

The acute triangle $ABC$ with $AB\neq AC$ has circumcircle $\Gamma$, circumcenter $O$, and orthocenter $H$. The midpoint of $BC$ is $M$, and the extension of the median $AM$ intersects $\Gamma$ at $N$. The circle of diameter $AM$ intersects $\Gamma$ again at $A$ and $P$. Show that the lines $AP$, $BC$, and $OH$ are concurrent if and only if $AH = HN$.

Two circles in the plane, $\gamma_1$ and $\gamma_2$, meet at points $M$ and $N$. Let $A$ be a point on $\gamma_1$, and let $D$ be a point on $\gamma_2$. The lines $AM$ and $AN$ meet again $\gamma_2$ at points $B$ and $C$, respectively, and the lines $DM$ and $DN$ meet again $\gamma_1$ at points $E$ and $F$, respectively. Assume the order $M$, $N$, $F$, $A$, $E$ is circular around $\gamma_1$, and the segments $AB$ and $DE$ are congruent. Prove that the points $A$, $F$, $C$ and $D$ lie on a circle whose centre does not depend on the position of the points $A$ and $D$ on the respective circles, subject to the assumptions above. [i]***[/i]

Consider an acute triangle $ABC$ of area $S$. Let $CD \perp AB$ ($D \in AB$), $DM \perp AC$ ($M \in AC$) and $DN \perp BC$ ($N \in BC$). Denote by $H_1$ and $H_2$ the orthocentres of the triangles $MNC$, respectively $MND$. Find the area of the quadrilateral $AH_1BH_2$ in terms of $S$.

Let points $A_1$, $A_2$ and $A_3$ lie on the circle $\Gamma$ in a counter-clockwise order, and let $P$ be a point in the same plane. For $i \in \{1,2,3\}$, let $\tau_i$ denote the counter-clockwise rotation of the plane centred at $A_i$, where the angle of rotation is equial to the angle at vertex $A_i$ in $\triangle A_1A_2A_3$. Further, define $P_i$ to be the point $\tau_{i+2}(\tau_{i}(\tau_{i+1}(P)))$, where the indices are taken modulo $3$ (i.e., $\tau_4 = \tau_1$ and $\tau_5 = \tau_2$). Prove that the radius of the circumcircle of $\triangle P_1P_2P_3$ is at most the radius of $\Gamma$. [i]Proposed by Anant Mudgal[/i]

Let $ABCD$ be a parallelogram whose diagonals meet at $P$. Denote by $M$ the midpoint of $AB$. Let $Q$ be a point such that $QA$ is tangent to the circumcircle of $MAD$ and $QB$ is tangent to the circumcircle of $MBC$. Prove that points $Q,M,P$ are collinear. (Patrik Bak, Slovakia)

Given a scalene triangle $\vartriangle ABC$ with orthocenter $H$ and circumcenter $O$. The exterior angle bisector of $\angle BAC$ intersects circumcircle of $\vartriangle ABC$ at $N \ne A$. Let $D$ be another intersection of $HN$ and the circumcircle of $\vartriangle ABC$. The line passing through $O$, which is parallel to $AN$, intersects $AB,AC$ at $E, F$, respectively. Prove that $DH$ bisects the angle $\angle EDF$. [img]https://3.bp.blogspot.com/-F1mFwojG_I0/XnYNR8ofqSI/AAAAAAAALeo/zge24WF0EO8umPAaXprKAeXJHAj7pr6tQCK4BGAYYCw/s1600/imoc2019g5.png[/img]

Let $ \triangle ABC$ be a right-angled triangle with $ \angle C\equal{}90^\circ$. $ CD$ is the altitude from $ C$ to $ AB$, with $ D$ on $ AB$. $ \omega$ is the circumcircle of $ \triangle BCD$. $ \omega_1$ is a circle situated in $ \triangle ACD$, it is tangent to the segments $ AD$ and $ AC$ at $ M$ and $ N$ respectively, and is also tangent to circle $ \omega$. (i) Show that $ BD\cdot CN\plus{}BC\cdot DM\equal{}CD\cdot BM$. (ii) Show that $ BM\equal{}BC$.

In triangle $ABC$, $AB>AC$. The bisectors of $\angle{B},\angle{C}$ intersect the sides $AC,AB$ at $P,Q$, respectively. Let $I$ be the incenter of $\Delta ABC$. Suppose that $IP=IQ$. How much isthe value of $\angle A$?

In triangle $ABC$ the midpoint of $BC$ is called $M$. Let $P$ be a variable interior point of the triangle such that $\angle CPM=\angle PAB$. Let $\Gamma$ be the circumcircle of triangle $ABP$. The line $MP$ intersects $\Gamma$ a second time at $Q$. Define $R$ as the reflection of $P$ in the tangent to $\Gamma$ at $B$. Prove that the length $|QR|$ is independent of the position of $P$ inside the triangle.

Let $ABC$ be a triangle with inradius $r$, circumradius $R$, and with sides $a=BC,b=CA,c=AB$. Prove that \[\frac{R}{2r} \ge \left(\frac{64a^2b^2c^2}{(4a^2-(b-c)^2)(4b^2-(c-a)^2)(4c^2-(a-b)^2)}\right)^2.\]