In isosceles trapezoid $ABCD$ ($AB \parallel CD$) points $E$ and $F$ lie on the segment $CD$ in such a way that $D, E, F$ and $C$ are in that order and $DE = CF$. Let $X$ and $Y$ be the reflection of $E$ and $C$ with respect to $AD$ and $AF$. Prove that circumcircles of triangles $ADF$ and $BXY$ are concentric. [i]Proposed by Iman Maghsoudi - Iran[/i]

Let $\alpha, \beta, \gamma, \delta$ be the interior angles of a convex quadrilateral, If $$ \cos\alpha + \cos\beta + \cos\gamma, + \cos\delta = 0 , $$ then this quadrilateral is cyclic or a trapezium. Prove it.

Let $\omega$ be incircle of $ABC$. $P$ and $Q$ are on $AB$ and $AC$, such that $PQ$ is parallel to $BC$ and is tangent to $\omega$. $AB,AC$ touch $\omega$ at $F,E$. Prove that if $M$ is midpoint of $PQ$, and $T$ is intersection point of $EF$ and $BC$, then $TM$ is tangent to $\omega$. [i]By Ali Khezeli[/i]

In the unit circle shown in the figure, chords $PQ$ and $MN$ are parallel to the unit radius $OR$ of the circle with center at $O$. Chords $MP$, $PQ$, and $NR$ are each $s$ units long and chord $MN$ is $d$ units long. [asy] draw(Circle((0,0),10)); draw((0,0)--(10,0)--(8.5,5.3)--(-8.5,5.3)--(-3,9.5)--(3,9.5)); dot((0,0)); dot((10,0)); dot((8.5,5.3)); dot((-8.5,5.3)); dot((-3,9.5)); dot((3,9.5)); label("1", (5,0), S); label("s", (8,2.6)); label("d", (0,4)); label("s", (-5,7)); label("s", (0,8.5)); label("O", (0,0),W); label("R", (10,0), E); label("M", (-8.5,5.3), W); label("N", (8.5,5.3), E); label("P", (-3,9.5), NW); label("Q", (3,9.5), NE); [/asy] Of the three equations \[ \textbf{I.}\ d-s=1, \qquad \textbf{II.}\ ds=1, \qquad \textbf{III.}\ d^2-s^2=\sqrt{5} \]those which are necessarily true are $\textbf{(A)}\ \textbf{I}\ \text{only} \qquad\textbf{(B)}\ \textbf{II}\ \text{only} \qquad\textbf{(C)}\ \textbf{III}\ \text{only} \qquad\textbf{(D)}\ \textbf{I}\ \text{and}\ \textbf{II}\ \text{only} \qquad\textbf{(E)}\ \textbf{I, II}\ \text{and} \textbf{III}$

In trapezoid $ABCD$, $AD$ is perpendicular to $DC$, $AD=AB=3$, and $DC=6$. In addition, E is on $DC$, and $BE$ is parallel to $AD$. Find the area of $\Delta BEC$. [asy] defaultpen(linewidth(0.7)); pair A=(0,3), B=(3,3), C=(6,0), D=origin, E=(3,0); draw(E--B--C--D--A--B); draw(rightanglemark(A, D, C)); label("$A$", A, NW); label("$B$", B, NW); label("$C$", C, SE); label("$D$", D, SW); label("$E$", E, NW); label("$3$", A--D, W); label("$3$", A--B, N); label("$6$", E, S);[/asy] $\textbf{(A)} \: 3\qquad \textbf{(B)} \: 4.5\qquad \textbf{(C)} \: 6\qquad \textbf{(D)} \: 9\qquad \textbf{(E)} \: 18\qquad $

Let $O$ be the point of intersection of the diagonals of the trapezoid $ABCD$ with the bases $AB$ and $CD$. It is known that $\angle AOB = \angle DAB = 90^o$. On the sides $AD$ and $BC$ take the points $E$ and $F$ so that $EF\parallel AB$ and $EF = AD$. Find the angle $\angle AOE$.

The lateral sidelines $AB$ and $CD$ of trapezoid $ABCD$ meet at point $S$. The bisector of angle $ASC$ meets the bases of the trapezoid at points $K$ and $L$ ($K$ lies inside segment $SL$). Point $X$ is chosen on segment $SK$, and point $Y$ is selected on the extension of $SL$ beyond $L$ such a way that $\angle AXC - \angle AYC = \angle ASC$. Prove that $\angle BXD - \angle BYD = \angle BSD$.

Given an isosceles trapezoid $ABCD$ and a point $P$. Prove that a quadrilateral can be constructed from segments $PA, PB, PC, PD$. Note: It is allowed that the vertices of a quadrilateral lie not only not only on the sides of the trapezoid, but also on their extensions.

In the figure, $ABCD$ is an isosceles trapezoid with side lengths $AD = BC = 5, AB = 4,$ and $DC = 10$. The point $C$ is on $\overline{DF}$ and $B$ is the midpoint of hypotenuse $\overline{DE}$ in the right triangle $DEF$. Then $CF =$ [asy] size(200); defaultpen(fontsize(10)); pair D=origin, A=(3,4), B=(7,4), C=(10,0), E=(14,8), F=(14,0); draw(B--C--F--E--B--A--D--B^^C--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F); pair point=(5,3); label("$A$", A, N); label("$B$", B, N); label("$C$", C, S); label("$D$", D, S); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); markscalefactor=0.05; draw(rightanglemark(E,F,D), linewidth(0.7));[/asy] $\text{(A)} \ 3.25 \qquad \text{(B)} \ 3.5 \qquad \text{(C)} \ 3.75 \qquad \text{(D)} \ 4.0 \qquad \text{(E)} \ 4.25$

Let $ABCD$ be a trapezium such that $AB||CD$ and $AB+CD=AD$. Let $P$ be the point on $AD$ such that $AP=AB$ and $PD=CD$. $a)$ Prove that $\angle BPC=90^{\circ}$. $b)$ $Q$ is the midpoint of $BC$ and $R$ is the point of intersection between the line $AD$ and the circle passing through the points $B,A$ and $Q$. Show that the points $B,P,R$ and $C$ are concyclic.

In triangle $ABC$ with $AB \neq AC$, points $N \in CA$, $M \in AB$, $P \in BC$, and $Q \in BC$ are chosen such that $MP \parallel AC$, $NQ \parallel AB$, $\frac{BP}{AB} = \frac{CQ}{AC}$, and $A, M, Q, P, N$ are concyclic. Find $\angle BAC$.

Let $ABCD$ be a trapezoid such that $AB \parallel CD$, $|BC|+|AD| = 7$, $|AB| = 9$ and $|BC| = 14$. What is the ratio of the area of the triangle formed by $CD$, angle bisector of $\widehat{BCD}$ and angle bisector of $\widehat{CDA}$ over the area of the trapezoid? $ \textbf{a)}\ \dfrac{9}{14} \qquad\textbf{b)}\ \dfrac{5}{7} \qquad\textbf{c)}\ \sqrt 2 \qquad\textbf{d)}\ \dfrac{49}{69} \qquad\textbf{e)}\ \dfrac 13 $

Let $P$ be a variable point on arc $BC$ of the circumcircle of triangle $ABC$ not containing $A$. Let $I_1$ and $I_2$ be the incenters of the triangles $PAB$ and $PAC$, respectively. Prove that: [b](a)[/b] The circumcircle of $?PI_1I_2$ passes through a fixed point. [b](b)[/b] The circle with diameter $I_1I_2$ passes through a fixed point. [b](c)[/b] The midpoint of $I_1I_2$ lies on a fixed circle.

Let $ABC$ be a triangle such that $AB<AC$. Let $D$ be a point on the segment $BC$ such that $BD<CD$. The angle bisectors of $\angle ADB$ and $\angle ADC$ meet the segments $AB$ and $AC$ at $E$ and $F$ respectively. Let $\omega$ be the circumcircle of $AEF$ and $M$ be the midpoint of $EF$. The ray $AD$ meets $\omega$ at $X$ and the line through $X$ parallel to $EF$ meets $\omega$ again at $Y$. If $YM$ meets $\omega$ at $T$, show that $AT$, $EF$ and $BC$ are concurrent. [i]Authored by Nikola Velov[/i]

Let $\omega$ be the circumcircle of a triangle $ABC$. Denote by $M$ and $N$ the midpoints of the sides $AB$ and $AC$, respectively, and denote by $T$ the midpoint of the arc $BC$ of $\omega$ not containing $A$. The circumcircles of the triangles $AMT$ and $ANT$ intersect the perpendicular bisectors of $AC$ and $AB$ at points $X$ and $Y$, respectively; assume that $X$ and $Y$ lie inside the triangle $ABC$. The lines $MN$ and $XY$ intersect at $K$. Prove that $KA=KT$.