Let $ABCD$ be a convex cyclic quadrilateral with diagonals $AC$ and $BD$. Each of the four vertixes are reflected across the diagonal on which the do not lie. (a) Investigate when the four points thus obtained lie on a straight line and give as simple an equivalent condition as possible to the cyclic quadrilateral $ABCD$ for it. (b) Show that in all other cases the four points thus obtained lie on one circle. (Theresia Eisenkölbl)

The diagonals of the cyclic quadrilateral $ABCD$ intersect at the point $E$. Let $P$ and $Q$ are the centers of the circles circumscribed around the triangles $BCE$ and $DCE$, respectively. A straight line passing through the point $P$ parallel to $AB$, and a straight line passing through the point $Q$ parallel to $AD$, intersect at the point $R$. Prove that the point $R$ lies on segment $AC$.

Given a circle with four circles that intersect in pairs as shown in the figure. The "internal" the points of intersection are $A, B, C$ and $D$, while the ‘outer’ points of intersection are $E, F, G$ and $H$. Prove that the quadrilateral $ABCD$ is cyclic if also the quadrilateral $EFGH$ is also cyclic. [img]https://cdn.artofproblemsolving.com/attachments/0/0/6a369c93e37eefd57775fd8586bdff393e1914.png[/img]

Let $\triangle ABC$ be a right-angled triangle with $\angle BAC = 90^{\circ}$ and let $E$ be the foot of the perpendicular from $A$ to $BC$. Let $Z \ne A$ be a point on the line $AB$ with $AB = BZ$. Let $(c)$ be the circumcircle of the triangle $\triangle AEZ$. Let $D$ be the second point of intersection of $(c)$ with $ZC$ and let $F$ be the antidiametric point of $D$ with respect to $(c)$. Let $P$ be the point of intersection of the lines $FE$ and $CZ$. If the tangent to $(c)$ at $Z$ meets $PA$ at $T$, prove that the points $T$, $E$, $B$, $Z$ are concyclic. Proposed by [i]Theoklitos Parayiou, Cyprus[/i]

Prove that the product of the sides of a quadrilateral inscribed in a circle with radius $1$ does not exceed $4$.

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]

The diagonals of a cyclic quadrilateral $ABCD$ meet at $O$. Let $S_1, S_2$ be the circumcircles of triangles $ABO$ and $CDO$ respectively, and $O,K$ their intersection points. The lines through $O$ parallel to $AB$ and $CD$ meet $S_1$ and $S_2$ again at $L$ and $M$, respectively. Points $P$ and $Q$ on segments $OL$ and $OM$ respectively are taken such that $OP : PL = MQ : QO$. Prove that $O,K, P,Q$ lie on a circle.

In the acute-angled triangle $ABC$, the point $F$ is the foot of the altitude from $A$, and $P$ is a point on the segment $AF$. The lines through $P$ parallel to $AC$ and $AB$ meet $BC$ at $D$ and $E$, respectively. Points $X \ne A$ and $Y \ne A$ lie on the circles $ABD$ and $ACE$, respectively, such that $DA = DX$ and $EA = EY$. Prove that $B, C, X,$ and $Y$ are concyclic.

Let $E$ be the point of intersection of the diagonals of the cyclic quadrilateral $ABCD$, and let $K, L, M$ and $N$ be the midpoints of the sides $AB, BC, CD$ and $DA$, respectively. Prove that the radii of the circles circumscribed around the triangles $KLE$ and $MNE$ are equal.

Let $ABC$ be an arbitrary triangle. A circle passes through $B$ and $C$ and intersects the lines $AB$ and $AC$ at $D$ and $E$, respectively. The projections of the points $B$ and $E$ on $CD$ are denoted by $B'$ and $E'$, respectively. The projections of the points $D$ and $C$ on $BE$ are denoted by $D'$ and $C'$, respectively. Prove that the points $B',D',E'$ and $C'$ lie on the same circle.

Given a cyclic quadrilateral $ABCD$. There is a point $P$ on side $BC$ such that $\angle PAB=\angle PDC=90^\circ$. The medians of vertexes $A$ and $D$ in triangles $PAB$ and $PDC$ meet at $K$ and the bisectors of $\angle PAB$ and $\angle PDC$ meet at $L$. Prove that $KL\perp BC$.

Let $ABCD$ be a cyclic quadrilateral which is not a trapezoid and whose diagonals meet at $E$. The midpoints of $AB$ and $CD$ are $F$ and $G$ respectively, and $\ell$ is the line through $G$ parallel to $AB$. The feet of the perpendiculars from E onto the lines $\ell$ and $CD$ are $H$ and $K$, respectively. Prove that the lines $EF$ and $HK$ are perpendicular.

In a triangle $ABC$, let $D$ be the point on the segment $BC$ such that $AB+BD=AC+CD$. Suppose that the points $B$, $C$ and the centroids of triangles $ABD$ and $ACD$ lie on a circle. Prove that $AB=AC$.

$ABCD$ is a cyclic quadrilateral with circumcenter $O$ and circumradius $7$. $AB$ intersects $CD$ at $E$, $DA$ intersects $CB$ at $F$. $OE=13$, $OF=14$. Let $\cos\angle FOE=\dfrac pq$, with $p$, $q$ coprime. Find $p+q$.

Let $ABCD$ be a cyclic quadrilateral such that $AB = AD + BC$ and $CD < AB$. The diagonals $AC$ and $BD$ intersect at $P$, while the lines $AD$ and $BC$ intersect at $Q$. The angle bisector of $\angle APB$ meets $AB$ at $T$. Show that the circumcenter of the triangle $CTD$ lies on the circumcircle of the triangle $CQD$. [i]Proposed by Nikola Velov[/i]

let the incircle of a triangle ABC touch BC,AC,AB at A1,B1,C1 respectively. M and N are the midpoints of AB1 and AC1 respectively. MN meets A1C1 at T . draw two tangents TP and TQ through T to incircle. PQ meets MN at L and B1C1 meets PQ at K . assume I is the center of the incircle . prove IK is parallel to AL

We are given a cyclic quadrilateral $ABCD$ with a point $E$ on the diagonal $AC$ such that $AD=AE$ and $CB=CE$. Let $M$ be the center of the circumcircle $k$ of the triangle $BDE$. The circle $k$ intersects the line $AC$ in the points $E$ and $F$. Prove that the lines $FM$, $AD$ and $BC$ meet at one point. [i](4th Middle European Mathematical Olympiad, Individual Competition, Problem 3)[/i]

Given a rhombus $ABCD$, find the locus of the points $P$ lying inside the rhombus and satisfying $\angle APD+\angle BPC=180^{\circ}$.

Let $ABC$ be a triangle with incentre $I$. The angle bisectors $AI$, $BI$ and $CI$ meet $[BC]$, $[CA]$ and $[AB]$ at $D$, $E$ and $F$, respectively. The perpendicular bisector of $[AD]$ intersects the lines $BI$ and $CI$ at $M$ and $N$, respectively. Show that $A$, $I$, $M$ and $N$ lie on a circle.

Let $ABCD$ be a cyclic quadrilateral such that $AB \cdot BC=2 \cdot AD \cdot DC$. Prove that its diagonals $AC$ and $BD$ satisfy the inequality $8BD^2 \le 9AC^2$. [color=#FF0000]Moderator says: Use the search before posting contest problems [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=530783[/url][/color]

Let $ABCD$ be any convex quadrilateral and let $P, Q, R, S$ be points on the segments $AB, BC, CD$, and $DA$, respectively. It is given that the segments $PR$ and $QS$ dissect $ABCD$ into four quadrilaterals, each of which has perpendicular diagonals. Show that the points $P, Q, R, S$ are concyclic.

Let $ABCD$ be a cyclic quadrilateral inscribed in circle $\omega$ whose diagonals meet at $F$. Lines $AB$ and $CD$ meet at $E$. Segment $EF$ intersects $\omega$ at $X$. Lines $BX$ and $CD$ meet at $M$, and lines $CX$ and $AB$ meet at $N$. Prove that $MN$ and $BC$ concur with the tangent to $\omega$ at $X$. [i]Proposed by Allen Liu[/i]

Let $ABCD$ be a cyclic quadrilateral with $AB$ and $CD$ not parallel. Let $M$ be the midpoint of $CD.$ Let $P$ be a point inside $ABCD$ such that $P A = P B = CM.$ Prove that $AB, CD$ and the perpendicular bisector of $MP$ are concurrent.

Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.

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 $CD\cdot EF+DF\cdot AE=BD\cdot AF$.