$ABCD$ is a trapezoid with $AB || CD$. There are two circles $\omega_1$ and $\omega_2$ is the trapezoid such that $\omega_1$ is tangent to $DA$, $AB$, $BC$ and $\omega_2$ is tangent to $BC$, $CD$, $DA$. Let $l_1$ be a line passing through $A$ and tangent to $\omega_2$(other than $AD$), Let $l_2$ be a line passing through $C$ and tangent to $\omega_1$ (other than $CB$). Prove that $l_1 || l_2$.

Let $ABCD$ be a trapezium and $AB$ and $CD$ be it's parallel edges. Find, with proof, the set of interior points $P$ of the trapezium which have the property that $P$ belongs to at least two lines each intersecting the segments $AB$ and $CD$ and each dividing the trapezium in two other trapezoids with equal areas.

Let $A,B$ and $C$ be three points on a line with $B$ between $A$ and $C$. Let $\Gamma_1,\Gamma_2, \Gamma_3$ be semicircles, all on the same side of $AC$ and with $AC,AB,BC$ as diameters, respectively. Let $l$ be the line perpendicular to $AC$ through $B$. Let $\Gamma$ be the circle which is tangent to the line $l$, tangent to $\Gamma_1$ internally, and tangent to $\Gamma_3$ externally. Let $D$ be the point of contact of $\Gamma$ and $\Gamma_3$. The diameter of $\Gamma$ through $D$ meets $l$ in $E$. Show that $AB=DE$.

$ABCD$ is a trapezium where $AD\parallel BC$. $P$ is the point on the line $AB$ such that $\angle CPD$ is maximal. $Q$ is the point on the line $CD$ such that $\angle BQA$ is maximal. Given that $P$ lies on the segment $AB$, prove that $\angle CPD=\angle BQA$.

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 $\angle BAD+\angle ADC = 120$.

$BC$ is a diameter of a circle and the points $X,Y$ are on the circle such that $XY\perp BC$. The points $P,M$ are on $XY,CY$ (or their stretches), respectively, such that $CY||PB$ and $CX||PM$. Let $K$ be the meet point of the lines $XC,BP$. Prove that $PB\perp MK$.

Let $ABCD$ be a square of side length $4$. Points $E$ and $F$ are chosen on sides $BC$ and $DA$, respectively, such that $EF = 5$. Find the sum of the minimum and maximum possible areas of trapezoid $BEDF$. [i]Proposed by Andrew Wu[/i]

Let $ABCD$ be an isosceles trapezoid, whose dimensions are $AB = 6$, $BC=5=DA$, and $CD=4$. Draw circles of radius 3 centered at $A$ and $B$, and circles of radius 2 centered at $C$ and $D$. A circle contained within the trapezoid is tangent to all four of these circles. Its radius is $\frac{-k+m\sqrt{n}}p$, where $k$, $m$, $n$, and $p$ are positive integers, $n$ is not divisible by the square of any prime, and $k$ and $p$ are relatively prime. Find $k+m+n+p$.

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 $ABCD$ be a trapezoid with $AB // CD$, $2|AB| = |CD|$ and $BD \perp BC$. Let $M$ be the midpoint of $CD$ and let $E$ be the intersection $BC$ and $AD$. Let $O$ be the intersection of $AM$ and $BD$. Let $N$ be the intersection of $OE$ and $AB$. (a) Prove that $ABMD$ is a rhombus. (b) Prove that the line $DN$ passes through the midpoint of the line segment $BE$.

$\triangle{ABC}$ is equailateral. $E$ is any point on $\overline{AC}$ produced and the equilateral $\triangle{ECD}$ is drawn. If $M$ and $N$ are the midpoints of $\overline{AD}$ and $\overline{EB}$ respectively then show that $\triangle{CMN}$ is equilateral.

In a trapezoid $ABCD$ with $AB<CD$ and $AB \parallel CD$, the diagonals intersect each other at $E$. Let $F$ be the midpoint of the arc $BC$ (not containing the point $E$) of the circumcircle of the triangle $EBC$. The lines $EF$ and $BC$ intersect at $G$. The circumcircle of the triangle $BFD$ intersects the ray $[DA$ at $H$ such that $A \in [HD]$. The circumcircle of the triangle $AHB$ intersects the lines $AC$ and $BD$ at $M$ and $N$, respectively. $BM$ intersects $GH$ at $P$, $GN$ intersects $AC$ at $Q$. Prove that the points $P, Q, D$ are collinear.

Let $ABCD$ be a trapezoid inscribed in a unit circle with diameter $AB$. If $DC = 4AD$, compute $AD$.

Given a convex, $n$-sided polygon $P$, form a $2n$-sided polygon $\text{clip}(P)$ by cutting off each corner of $P$ at the edges' trisection points. In other words, $\text{clip}(P)$ is the polygon whose vertices are the $2n$ edge trisection points of $P$, connected in order around the boundary of $P$. Let $P_1$ be an isosceles trapezoid with side lengths $13,13,13,$ and $3$, and for each $i\geq 2$, let $P_i=\text{clip}(P_{i-1}).$ This iterative clipping process approaches a limiting shape $P_\infty=\lim_{i\to\infty}P_i$. If the difference of the areas of $P_{10}$ and $P_\infty$ is written as a fraction $\tfrac xy$ in lowest terms, calculate the number of positive integer factors of $x\cdot y$.

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.

In a trapezoid $ABCD$, the internal bisector of angle $A$ intersects the base $BC$(or its extension) at the point $E$. Inscribed in the triangle $ABE$ is a circle touching the side $AB$ at $M$ and side $BE$ at the point $P$. Find the angle $DAE$ in degrees, if $AB:MP=2$.

In a trapezoid, the length of one of the diagonals is equal to the sum of the lengths of the bases, and the angle between the diagonals is $60$ degrees. Prove that this trapezoid is isosceles.

Given that points $ D,E$ lie on the sidelines $ AB,BC$ of triangle $ ABC$, respectively, point $ P$ is in interior of triangle $ ABC$ such that $ PE \equal{} PC$ and $ \bigtriangleup DEP\sim \bigtriangleup PCA.$ Prove that $ BP$ is tangent of the circumcircle of triangle $ PAD.$

A trapezoid $ABCD$ ($AB || CF$,$AB > CD$) is circumscribed.The incircle of the triangle $ABC$ touches the lines $AB$ and $AC$ at the points $M$ and $N$,respectively.Prove that the incenter of the trapezoid $ABCD$ lies on the line $MN$.