Let $ABCD$ be a isosceles trapezoid. Base $AD$ is $11$ cm long while the other three sides are each $5$ cm long. We draw the line that is perpendicular to $BD$ and contains $C$ and the line that is perpendicular to $AC$ and contains$ B$. We mark the intersection of these two lines with $E$. What is the distance between point $E$ and line $AD$?

Let $ABC$ be a scalene triangle with circumcircle $\Gamma$, and let $D$,$E$,$F$ be the points where its incircle meets $BC$, $AC$, $AB$ respectively. Let the circumcircles of $\triangle AEF$, $\triangle BFD$, and $\triangle CDE$ meet $\Gamma$ a second time at $X,Y,Z$ respectively. Prove that the perpendiculars from $A,B,C$ to $AX,BY,CZ$ respectively are concurrent. [i]Proposed by Michael Kural[/i]

The side lengths of a trapezoid are $\sqrt[4]{3}, \sqrt[4]{3}, \sqrt[4]{3}$, and $2 \cdot \sqrt[4]{3}$. Its area is the ratio of two relatively prime positive integers, $m$ and $n$. Find $m + n$.

Let $\Omega$ be the circumcircle of an isosceles trapezoid $ABCD$, in which $AD$ is parallel to $BC$. Let $X$ be the reflection point of $D$ with respect to $BC$. Point $Q$ is on the arc $BC$ of $\Omega$ that does not contain $A$. Let $P$ be the intersection of $DQ$ and $BC$. A point $E$ satisfies that $EQ$ is parallel to $PX$, and $EQ$ bisects $\angle BEC$. Prove that $EQ$ also bisects $\angle AEP$. [i]Proposed by Li4.[/i]

The internal angles of quadrilateral $ABCD$ form an arithmetic progression. Triangles $ABD$ and $DCB$ are similar with $\angle DBA=\angle DCB$ and $\angle ADB=\angle CBD$. Moreover, the angles in each of these two triangles also form an arithmetic progression. In degrees, what is the largest possible sum of the two largest angles of $ABCD$? ${\textbf{(A)}\ 210\qquad\textbf{(B)}\ 220\qquad\textbf{(C)}\ 230\qquad\textbf{(D}}\ 240\qquad\textbf{(E)}\ 250$

Inside the square $ABCD$ mark the point $P$, for which $\angle BAP = 30^o$ and $\angle BCP = 15^o$. The point $Q$ was chosen so that $APCQ$ is an isosceles trapezoid ($PC\parallel AQ$). Find the angles of the triangle $CAM$, where $M$ is the midpoint of $PQ$.

Consider a trapezoid $ABCD$ with $BC||AD$ and $BC<AD$. Let the lines $AB$ and $CD$ meet at $X$. Let $\omega_1$ be the incircle of the triangle $XBC$, and let $\omega_2$ be the excircle of the triangle $XAD$ which is tangent to the segment $AD$ . Denote by $a$ and $d$ the lines tangent to $\omega_1$ , distinct from $AB$ and $CD$, and passing through $A$ and $D$, respectively. Denote by $b$ and $c$ the lines tangent to $\omega_2$ , distinct from $AB$ and $CD$, passing through $B$ and $C$ respectively. Assume that the lines $a,b,c$ and $d$ are distinct. Prove that they form a parallelogram.

Let $ ABCD$ be a cyclic quadrilateral with $ AB>CD$ and $ BC>AD$. Take points $ X$ and $ Y$ on the sides $ AB$ and $ BC$, respectively, so that $ AX\equal{}CD$ and $ AD\equal{}CY$. Let $ M$ be the midpoint of $ XY$. Prove that $ AMC$ is a right angle.

Let $ABCD$ be a trapezoid with bases $AB$ and $CD$, inscribed in a circle of center $O$. Let $P$ be the intersection of the lines $BC$ and $AD$. A circle through $O$ and $P$ intersects the segments $BC$ and $AD$ at interior points $F$ and $G$, respectively. Show that $BF=DG$.

$ 5 $ points are situated in the plane so that any three of them form a triangle of area at most $ 1. $ Prove that there is a trapezoid of area at most $ 3 $ which contains all these points ('including' here means that the points can also be on the sides of the trapezoid).

There is a stone at each vertex of a given regular $13$-gon, and the color of each stone is black or white. Prove that we may exchange the position of two stones such that the coloring of these stones are symmetric with respect to some symmetric axis of the $13$-gon.

$ ABCD$ is convex $ AD\parallel BC$, $ AC\perp BD$. $ M$ is interior point of $ ABCD$ which is not a intersection of diagonals $ AC$ and $ BD$ such that $ \angle AMB \equal{}\angle CMD \equal{}\frac{\pi}{2}$ .$ P$ is intersection of angel bisectors of $ \angle A$ and $ \angle C$. $ Q$ is intersection of angel bisectors of $ \angle B$ and $ \angle D$. Prove that $ \angle PMB \equal{}\angle QMC$.

Let be an isosceles trapezoid such that its smaller base is equal to its legs, and a rhombus that has each of its vertexes on a different side of the trapezoid. Prove that the smaller angles of the trapezoid are equal to the smaller ones of the rhombus. [i]Vlad Robu[/i]

A (convex) trapezoid $ABCD$ is good, if it is inscribed in a circle, sides $AB$ and $CD$ are the bases and $CD$ is shorter than $AB$. For a good trapezoid $ABCD$ the following terms are defined: $\bullet$ The parallel to $AD$ passing through $B$ intersects the extension of side $CD$ at point $S$. $\bullet$ The two tangents passing through $S$ on the circumircle of the trapezoid touch the circle at $E$ and $F$, where $E$ lies on the same side of the straight line $CD$ as $A$. Give the simplest possible equivalent condition (expressed in side lengths and / or angles of the trapezoid) so that with a good trapezoid $ABCD$ the two angles $\angle BSE$ and $\angle FSC$ have the same measure. (Walther Janous)

Consider a non-zero convex angle $\angle POQ$ and its inner point $M$. Moreover, let $m>0$ be given. Construct a trapezoid $ABCD$ satisfying the following conditions: (1) vertices $A, D$ lie on ray $OP$ and vertices $B,C$ lie on ray $OQ$, (2) diagonals $AC$ and $BD$ intersect in $M$, (3) $AB=m$. Prove that your construction is correct and discuss conditions of solvability.