Let $ABCD$ be a trapezium with $AC = BC$. Let $H$ be the midpoint of the base $AB$ and let $\ell$ be a line passing through $H$. Let $\ell$ meet $AD$ at $P$ and $BD$ at $Q$. Prove that the angles $ACP$ and $QCB$ are either equal or have a sum of $180^o$. (I. Sharygin, Moscow)

Let $ABCD$ be a trapezoid with $AD\parallel BC$. A point $M $ is chosen inside the trapezoid, and a point $N$ is chosen inside the triangle $BMC$ such that $AM\parallel CN$, $BM\parallel DN$. Prove that triangles $ABN$ and $CDM$ have equal areas.

On the midline of the isosceles trapezoid $ABCD$ ($BC \parallel AD$) find the point $K$, for which the sum of the angles $\angle DAK + \angle BCK$ will be the smallest.

The diagonals of the trapezoid $ABCD$ are perpendicular ($AD//BC, AD>BC$) . Point $M$ is the midpoint of the side of $AB$, the point $N$ is symmetric of the center of the circumscribed circle of the triangle $ABD$ wrt $AD$. Prove that $\angle CMN = 90^o$. (A. Mudgal, India)

Points $E$ and $F$ are chosen on the base side $AD$ and the lateral side $AB$ of an isosceles trapezoid $ABCD$, respectively. Quadrilateral $CDEF$ is an isosceles trapezoid as well. Prove that $AE \cdot ED = AF \cdot FB$.

Given triangle $ABC$ and a point $P$ inside it, $\angle BAP=18^\circ$, $\angle CAP=30^\circ$, $\angle ACP=48^\circ$, and $AP=BC$. If $\angle BCP=x^\circ$, find $x$.

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.

Prove the following inequality. \[ \frac{2}{2\plus{}e^{\frac 12}}<\int_0^1 \frac{dx}{1\plus{}xe^{x}}<\frac{2\plus{}e}{2(1\plus{}e)}\]

Given points $A$, $B$, $C$, and $D$ on circle $\omega$ such that lines $AB$ and $CD$ intersect on point $T$ where $A$ is between $B$ and $T$, moreover $D$ is between $C$ and $T$. It is known that the line passing through $D$ which is parallel to $AB$ intersects $\omega$ again on point $E$ and line $ET$ intersects $\omega$ again on point $F$. Let $CF$ and $AB$ intersect on point $G$, $X$ be the midpoint of segment $AB$, and $Y$ be the reflection of point $T$ to $G$. Prove that $X$, $Y$, $C$, and $D$ are concyclic.

In the given circle, the diameter $\overline{EB}$ is parallel to $\overline{DC}$, and $\overline{AB}$ is parallel to $\overline{ED}$. The angles $AEB$ and $ABE$ are in the ratio $4:5$. What is the degree measure of angle $BCD$? [asy] unitsize(7mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real r=3; pair A=(-3cos(80),-3sin(80)); pair D=(3cos(80),3sin(80)), C=(-3cos(80),3sin(80)); pair O=(0,0), E=(-3,0), B=(3,0); path outer=Circle(O,r); draw(outer); draw(E--B); draw(E--A); draw(B--A); draw(E--D); draw(C--D); draw(B--C); pair[] ps={A,B,C,D,E,O}; dot(ps); label("$A$",A,N); label("$B$",B,NE); label("$C$",C,S); label("$D$",D,S); label("$E$",E,NW); label("$$",O,N);[/asy] $ \textbf{(A)}\ 120 \qquad \textbf{(B)}\ 125 \qquad \textbf{(C)}\ 130 \qquad \textbf{(D)}\ 135 \qquad \textbf{(E)}\ 140 $

Construct a trapezoid given the midpoints of the legs, the point of intersection of the diagonals and the foot of the perpendicular, drawn from this point on the larger base.

The sum of the angles on the bigger base of a trapezoid is $90^o$. Prove that the line segment whose ends are the midpoints of the bases, is equal to the line segment whose ends are the midpoints of the diagonals.

In trapezoid $ABCD$ with $AB || CD$, $\omega_1$ and $\omega_2$ are two circles with diameters $AD$ and $BC$, respectively. Let $X$ and $Y$ be two arbitrary points on $\omega_1$ and $\omega_2$, respectively. Show that the length of segment $XY$ is not more than half the perimeter of $ABCD$. [i]Proposed by Mahdi Etesami Fard[/i]

One base of a trapezoid is 100 units longer than the other base. The segment that joins the midpoints of the legs divides the trapezoid into two regions whose areas are in the ratio $2: 3.$ Let $x$ be the length of the segment joining the legs of the trapezoid that is parallel to the bases and that divides the trapezoid into two regions of equal area. Find the greatest integer that does not exceed $x^2/100.$

All of the polygons in the figure below are regular. Prove that $ABCD$ is an isosceles trapezoid. [img]https://cdn.artofproblemsolving.com/attachments/e/a/3f4de32becf4a90bf0f0b002fb4d8e724e8844.png[/img] [i]Proposed by Mahdi Etesamifard - Iran[/i]

In trapezoid $ABCD$: $BC <AD, AB = CD, K$ is midpoint of $AD, M$ is midpoint of $CD, CH$ is height. Prove that lines $AM, CK$ and $BH$ intersect at one point.

Consider a trapezium $AB \Gamma \Delta$, where $A\Delta \parallel B\Gamma$ and $\measuredangle A = 120^{\circ}$. Let $E$ be the midpoint of $AB$ and let $O_1$ and $O_2$ be the circumcenters of triangles $AE \Delta$ and $BE\Gamma$, respectively. Prove that the area of the trapezium is equal to six time the area of the triangle $O_1 E O_2$.

Given is a trapezoid $ ABCD$ where $ AB$ and $ CD$ are parallel, and $ A,B,C,D$ are clockwise in this order. Let $ \Gamma_1$ be the circle with center $ A$ passing through $ B$, $ \Gamma_2$ be the circle with center $ C$ passing through $ D$. The intersection of line $ BD$ and $ \Gamma_1$ is $ P$ $ ( \ne B,D)$. Denote by $ \Gamma$ the circle with diameter $ PD$, and let $ \Gamma$ and $ \Gamma_1$ meet at $ X$$ ( \ne P)$. $ \Gamma$ and $ \Gamma_2$ meet at $ Y$. If the circumcircle of triangle $ XBY$ and $ \Gamma_2$ meet at $ Q$, prove that $ B,D,Q$ are collinear.

In the plane we are given two circles intersecting at $ X$ and $ Y$. Prove that there exist four points with the following property: (P) For every circle touching the two given circles at $ A$ and $ B$, and meeting the line $ XY$ at $ C$ and $ D$, each of the lines $ AC$, $ AD$, $ BC$, $ BD$ passes through one of these points.

A trapezoid $ABCD$ with perpendicular diagonals $AC$ and $BD$ is inscribed in a circle $k$. Let $k_a$ and $k_c$ respectively be the circles with diameters $AB$ and $CD$. Compute the area of the region which is inside the circle $k$, but outside the circles $k_a$ and $k_c$.

Given a trapezoid $ABCD$ ($AD \parallel BC$) with $\angle ABC > 90^\circ$ . Point $M$ is chosen on the lateral side $AB$. Let $O_1$ and $O_2$ be the circumcenters of the triangles $MAD$ and $MBC$, respectively. The circumcircles of the triangles $MO_1D$ and $MO_2C$ meet again at the point $N$. Prove that the line $O_1O_2$ passes through the point $N$.

If the diagonals of a quadrilateral are perpendicular to each other, the figure would always be included under the general classification: $ \textbf{(A)}\ \text{rhombus} \qquad\textbf{(B)}\ \text{rectangles} \qquad\textbf{(C)}\ \text{square} \qquad\textbf{(D)}\ \text{isosceles trapezoid}\qquad\textbf{(E)}\ \text{none of these} $

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.

The diagonals of a trapezoid $ ABCD$ intersect at point $ P$. Point $ Q$ lies between the parallel lines $ BC$ and $ AD$ such that $ \angle AQD \equal{} \angle CQB$, and line $ CD$ separates points $ P$ and $ Q$. Prove that $ \angle BQP \equal{} \angle DAQ$. [i]Author: Vyacheslav Yasinskiy, Ukraine[/i]

$ABC$ is a right triangle with $\angle C=90$,$CD$ is perpendicular to $AB$,and $D$ is the foot,$\omega$ is the circumcircle of triangle $BCD$,$\omega_{1}$ is a circle inside triangle $ACD$,tangent to $AD$ and $AC$ at $M$ and $N$ respectively,and $\omega_{1}$ is also tangent to $\omega$.prove that: (1)$BD*CN+BC*DM=CD*BM$ (2)$BM=BC$