The cirumcentre of the cyclic quadrilateral $ABCD$ is $O$. The second intersection point of the circles $ABO$ and $CDO$, other than $O$, is $P$, which lies in the interior of the triangle $DAO$. Choose a point $Q$ on the extension of $OP$ beyond $P$, and a point $R$ on the extension of $OP$ beyond $O$. Prove that $\angle QAP=\angle OBR$ if and only if $\angle PDQ=\angle RCO$.

Refer to the diagram below. Let $ABCD$ be a cyclic quadrilateral with center $O$. Let the internal angle bisectors of $\angle A$ and $\angle C$ intersect at $I$ and let those of $\angle B$ and $\angle D$ intersect at $J$. Now extend $AB$ and $CD$ to intersect $IJ$ and $P$ and $R$ respectively and let $IJ$ intersect $BC$ and $DA$ at $Q$ and $S$ respectively. Let the midpoints of $PR$ and $QS$ be $M$ and $N$ respectively. Given that $O$ does not lie on the line $IJ$, show that $OM$ and $ON$ are perpendicular.

Let $ABC$ be an acute triangle with $AB<AC<BC$ and let $D$ be a point on it's extension of $BC$ towards $C$. Circle $c_1$, with center $A$ and radius $AD$, intersects lines $AC,AB$ and $CB$ at points $E,F$, and $G$ respectively. Circumscribed circle $c_2$ of triangle $AFG$ intersects again lines $FE,BC,GE$ and $DF$ at points $J,H,H' $ and $J'$ respectively. Circumscribed circle $c_3$ of triangle $ADE$ intersects again lines $FE,BC,GE$ and $DF$ at points $I,K,K' $ and $I' $ respectively. Prove that the quadrilaterals $HIJK$ and $H'I'J'K '$ are cyclic and the centers of their circumscribed circles coincide. by Evangelos Psychas, Greece

Let $ ABCD$ be a cyclic quadrilateral and $ O$ be the intersection of diagonal $ AC$ and $ BD$. The circumcircles of triangle $ ABO$ and the triangle $ CDO$ intersect at $ K$. Let $ L$ be a point such that the triangle $ BLC$ is similar to $ AKD$ (in that order). Prove that if $ BLCK$ is a convex quadrilateral, then it has an incircle.

In the quadrilateral $ABCD$ the angles $B{}$ and $D{}$ are straight. The lines $AB{}$ and $DC{}$ intersect at $E$ and the lines $AD$ and $BC$ intersect at $F{}.$ The line passing through $B{}$ parallel to $C{}$D intersects the circumscribed circle $\omega$ of $ABF{}$ at $K{}$ and the segment $KE{}$ intersects $\omega$ at $P{}.$ Prove that the line $AP$ divides the segment $CE$ in half.

David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, \(A\), \(B\), \(C\), which can each be inscribed in a circle with radius \(1\). Let \(\varphi_A\) denote the measure of the acute angle made by the diagonals of quadrilateral \(A\), and define \(\varphi_B\) and \(\varphi_C\) similarly. Suppose that \(\sin\varphi_A=\frac{2}{3}\), \(\sin\varphi_B=\frac{3}{5}\), and \(\sin\varphi_C=\frac{6}{7}\). All three quadrilaterals have the same area \(K\), which can be written in the form \(\frac{m}{n}\), where \(m\) and \(n\) are relatively prime positive integers. Find \(m+n\).

Points $A_1, B_1, C_1$ inside an acute-angled triangle $ABC$ are selected on the altitudes from $A, B, C$ respectively so that the sum of the areas of triangles $ABC_1, BCA_1$, and $CAB_1$ is equal to the area of triangle $ABC$. Prove that the circumcircle of triangle $A_1B_1C_1$ passes through the orthocenter $H$ of triangle $ABC$.

Let $ ABCD$ be a cyclic quadrilateral and $ r$ and $ s$ the lines obtained reflecting $ AB$ with respect to the internal bisectors of $ \angle CAD$ and $ \angle CBD$, respectively. If $ P$ is the intersection of $ r$ and $ s$ and $ O$ is the center of the circumscribed circle of $ ABCD$, prove that $ OP$ is perpendicular to $ CD$.

Let $ABCD$ be a cyclic quadrilateral. The diagonals $AC$ and $BD$ meet at $P$, and $DA $ and $CB$ meet at $Q$. Suppose $PQ$ is perpendicular to $AC$. Let $E$ be the midpoint of $AB$. Prove that $PE$ is perpendicular to $BC$.

Given a triangle $ABC$, let $D$ be the intersection of the line through $A$ perpendicular to $AB$, and the line through $B$ perpendicular to $BC$. Let $P$ be a point inside the triangle. Show that $DAPB$ is cyclic if and only if $\angle BAP = \angle CBP$.

A cyclic quadrilateral $ABCD$ is given. Let $I{}$ and $J{}$ be the centers of circles inscribed in the triangles $ABC$ and $ADC$. It turns out that the points $B, I, J, D$ lie on the same circle. Prove that the quadrilateral $ABCD$ is tangential.

Let $ABCD$ be an inscribed quadrilateral in a circle $c(O,R)$ (of circle $O$ and radius $R$). With centers the vertices $A,B,C,D$, we consider the circles $C_{A},C_{B},C_{C},C_{D}$ respectively, that do not intersect to each other . Circle $C_{A}$ intersects the sides of the quadrilateral at points $A_{1} , A_{2}$ , circle $C_{B}$ intersects the sides of the quadrilateral at points $B_{1} , B_{2}$ , circle $C_{C}$ at points $C_{1} , C_{2}$ and circle $C_{D}$ at points $C_{1} , C_{2}$ . Prove that the quadrilateral defined by lines $A_{1}A_{2} , B_{1}B_{2} , C_{1}C_{2} , D_{1}D_{2}$ is cyclic.

A cyclic quadrilateral $ABCD$ is given. Point $K_1, K_2$ lie on the segment $AB$, points $L_1, L_2$ on the segment $BC$, points $M_1, M_2$ on the segment $CD$ and points $N_1, N_2$ on the segment $DA$. Moreover, points $K_1, K_2, L_1, L_2, M_1, M_2, N_1, N_2$ lie on a circle $\omega$ in that order. Denote by $a, b, c, d$ the lengths of the arcs $N_2K_1, K_2L_1, L_2M_1, M _2N_1$ of the circle $\omega$ not containing points $K_2, L_2, M_2, N_2$, respectively. Prove that \begin{align*} a+c=b+d. \end{align*}

(A.Zaslavsky, 9--10) The reflections of diagonal $ BD$ of a quadrilateral $ ABCD$ in the bisectors of angles $ B$ and $ D$ pass through the midpoint of diagonal $ AC$. Prove that the reflections of diagonal $ AC$ in the bisectors of angles $ A$ and $ C$ pass through the midpoint of diagonal $ BD$ (There was an error in published condition of this problem).

Three circle of unit radius passing through the point $P$ and one of the points of $A, B$ and $C$ each. What can be the radius of the circumcircle of the triangle $ABC$?

The diagonals $AC$ and $BD$ of a cyclic quadrilateral $ABCD$ meet at $O$. The circumcircles of triangles $AOB$ and $COD$ intersect again at $K$. Point $L$ is such that the triangles $BLC$ and $AKD$ are similar and equally oriented. Prove that if the quadrilateral $BLCK$ is convex, then it is tangent [has an incircle].

Let $ABCD$ be a cyclic quadrilateral inscribed in a circle $\omega$ with center $O$. The lines $AD$ and $BC$ meet at $E$, while the lines $AB$ and $CD$ meet at $F$. Let $P$ be a point on the segment $EF$ such that $OP \perp EF$. The circle $\Gamma_{1}$ passes through $A$ and $E$ and is tangent to $\omega$ at $A$, while $\Gamma_{2}$ passes through $C$ and $F$ and is tangent to $\omega$ at $C$. If $\Gamma_{1}$ and $\Gamma_{2}$ meet at $X$ and $Y$, prove that $PO$ is the bisector of $\angle XPY$. [i]Proposed by Nikola Velov[/i]

On the diagonals $AC$ and $BD$ of the inscribed quadrilateral A$BCD$, the points $X$ and $Y$ are marked, respectively, so that the quadrilateral $ABXY$ is a parallelogram. Prove that the circumscribed circles of triangles $BXD$ and $CYA$ have equal radii. (Vyacheslav Yasinsky)

Denote by $ M$ midpoint of side $ BC$ in an isosceles triangle $ \triangle ABC$ with $ AC = AB$. Take a point $ X$ on a smaller arc $ \overarc{MA}$ of circumcircle of triangle $ \triangle ABM$. Denote by $ T$ point inside of angle $ BMA$ such that $ \angle TMX = 90$ and $ TX = BX$. Prove that $ \angle MTB - \angle CTM$ does not depend on choice of $ X$. [i]Author: Farzan Barekat, Canada[/i]

Let $ABC$ be a triangle such that $\angle C=2\angle B$ and $\omega$ be its circumcircle. a tangent from $A$ to $\omega$ intersect $BC$ at $E$. $\Omega$ is a circle passing throw $B$ that is tangent to $AC$ at $C$. Let $\Omega\cap AB=F$. $K$ is a point on $\Omega$ such that $EK$ is tangent to $\Omega$ ($A,K$ aren't in one side of $BC$). Let $M$ be the midpoint of arc $BC$ of $\omega$ (not containing $A$). Prove that $AFMK$ is a cyclic quadrilateral. [asy] import graph; size(15.424606256655986cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -7.905629294221492, xmax = 11.618976962434495, ymin = -5.154837585051625, ymax = 4.0091473316396895; /* image dimensions */ pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); /* draw figures */ draw(circle((1.4210145017438194,0.18096629151696939), 2.581514123077079)); draw(circle((1.4210145017438194,-1.3302878964546825), 2.8984706754484924)); draw(circle((-0.7076932767793396,-0.4161825262831505), 2.9101722408015513), linetype("4 4") + red); draw((3.996177869179178,0.)--(-3.839514259733819,0.)); draw((3.996177869179178,0.)--(0.07833180472267817,2.385828723227042)); draw((0.07833180472267817,2.385828723227042)--(-1.154148865691539,0.)); draw((-3.839514259733819,0.)--(-0.6807342461448075,-3.3262298939043657)); draw((0.07833180472267817,2.385828723227042)--(-3.839514259733819,0.)); /* dots and labels */ dot((3.996177869179178,0.),blue); label("$B$", (4.040279615036859,0.10218054796102663), NE * labelscalefactor,blue); dot((-1.154148865691539,0.),blue); label("$C$", (-1.3803811057738653,-0.14328333373606214), NE * labelscalefactor,blue); dot((1.4210145017438194,1.5681827789938092),linewidth(4.pt)); label("$F$", (1.4629088572174203,1.6465574703052102), NE * labelscalefactor); dot((0.07833180472267817,2.385828723227042),linewidth(3.pt) + blue); label("$A$", (-0.04055741817725232,2.5568193649319144), NE * labelscalefactor,blue); dot((-3.839514259733819,0.),linewidth(3.pt)); label("$E$", (-4.049800819229713,-0.06146203983703255), NE * labelscalefactor); dot((1.4210145017438194,-2.40054783156011),linewidth(4.pt) + uuuuuu); label("$M$", (1.4117705485305265,-2.6490604593938434), NE * labelscalefactor,uuuuuu); dot((-0.6807342461448075,-3.3262298939043657),linewidth(4.pt)); label("$K$", (-0.7871767250058992,-3.5490946922831688), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]

In right-angle triangle $ABC$, $\angle C=90$°, Point $D$ is the midpoint of side $AB$. Points $M$ and $C$ lie on the same side of $AB$ such that $MB\bot AB$, line $MD$ intersects side $AC$ at $N$, line $MC$ intersects side $AB$ at $E$. Show that $\angle DBN=\angle BCE$.

ABC is an isosceles triangle with AB=AC. Point D lies on side BC. Point F is inside $\triangle$ABC and lies on the circumcircle of triangle ADC. The circumcircle of triangle BDF intersects side AB at point E. Prove that $CD\cdot EF+DF\cdot AE=BD\cdot AF$.

The lengths of the four sides of an cyclic octagon are $4$ cm, the lengths of the other four sides are $6$ cm. Find the area of ​​the octagon.

Let $PQRS$ be a cyclic quadrilateral with $\angle PSR = 90^o$, and let $H,K$ be the projections of $Q$ on the lines $PR$ and $PS$, respectively. Prove that the line $HK$ passes through the midpoint of the segment $SQ$.

The quadrilateral $ABCD$ is inscribed in the circle and the lengths of the sides $BC$ and $DC$ are equal, and the length of the side $AB$ is equal to the length of the diagonal $AC$. Let the point $P$ be the midpoint of the arc $CD$, which does not contain point $A$, and $Q$ is the point of intersection of diagonals $AC$ and $BD$. Prove that the lines $PQ$ and $AB$ are perpendicular.