In the right-angled trapezoid $AB CD$, $AB \parallel CD$, $m( \angle A) = 90°$, $AD = DC = a$ and $AB =2a$. On the perpendiculars raised in $C$ and $D$ on the plane containing the trapezoid one considers points $E$ and $F$ (on the same side of the plane) such that $CE = 2a$ and $DF = a$. Find the distance from the point $B$ to the plane $(AEF)$ and the measure of the angle between the lines $AF$ and $BE$.

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.

Prove that in an isosceles trapezoid circumscibed around a circle, the segments connecting the points of tangency of opposite sides with the circle pass through the point of intersection of the diagonals.

Let $ABCD$ be a trapezoid in which $AB \parallel CD$ and $AB = 2CD$. A line $\ell$ perpendicular to $CD$ was drawn through point $C$. A circle with center at point $D$ and radius $DA$ intersects line $\ell$ at points $P$ and $Q$. Prove that $AP \perp BQ$.

Let $ ABCD$ be a trapezoid inscribed in a circle $ \mathcal{C}$ with $ AB\parallel CD$, $ AB\equal{}2CD$. Let $ \{Q\}\equal{}AD\cap BC$ and let $ P$ be the intersection of tangents to $ \mathcal{C}$ at $ B$ and $ D$. Calculate the area of the quadrilateral $ ABPQ$ in terms of the area of the triangle $ PDQ$.

In a trapezoid $ABCD$ whose parallel sides $AB$ and $CD$ are in ratio $\frac{AB}{CD}=\frac32$, the points $ N$ and $M$ are marked on the sides $BC$ and $AB$ respectively, in such a way that $BN = 3NC$ and $AM = 2MB$ and segments $AN$ and $DM$ are drawn that intersect at point $P$, find the ratio between the areas of triangle $APM$ and trapezoid $ABCD$. [img]https://cdn.artofproblemsolving.com/attachments/7/8/21d59ca995d638dfcb76f9508e439fd93a5468.png[/img]

Given a trapezoid, divide it by a line into two quadrilaterals in such a way that both of them are cyclic with the same circumradius. Discuss conditions of solvability.

Let $\Gamma$ and $\Gamma'$ be two congruent circles centered at $O$ and $O'$, respectively, and let $A$ be one of their two points of intersection. $B$ is a point on $\Gamma$, $C$ is the second point of intersection of $AB$ and $\Gamma'$, and $D$ is a point on $\Gamma'$ such that $OBDO'$ is a parallelogram. Show that the length of $CD$ does not depend on the position of $B$.

A rectangular piece of paper with side lengths 5 by 8 is folded along the dashed lines shown below, so that the folded flaps just touch at the corners as shown by the dotted lines. Find the area of the resulting trapezoid. [asy] size(150); defaultpen(linewidth(0.8)); draw(origin--(8,0)--(8,5)--(0,5)--cycle,linewidth(1)); draw(origin--(8/3,5)^^(16/3,5)--(8,0),linetype("4 4")); draw(origin--(4,3)--(8,0)^^(8/3,5)--(4,3)--(16/3,5),linetype("0 4")); label("$5$",(0,5/2),W); label("$8$",(4,0),S); [/asy]

In trapezoid $ABCD$, sides $AB$ and $CD$ are parallel, and diagonal $BD$ and side $AD$ have equal length. If $m\angle DBC=110^\circ$ and $m\angle CBD =30^\circ$, then $m \angle ADB=$ $\text{(A)}\ 80^\circ \qquad \text{(B)}\ 90^\circ \qquad \text{(C)}\ 100^\circ \qquad \text{(D)}\ 110^\circ \qquad \text{(E)}\ 120^\circ$

Let $ABCD$ be a quadrilateral with $\overline{AB}\parallel\overline{CD}$, $AB=16$, $CD=12$, and $BC<AD$. A circle with diameter $12$ is inside of $ABCD$ and tangent to all four sides. Find $BC$.

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.

An isosceles trapezoid $ABCD$ with bases $AB$ and $CD$ has $AB=13$, $CD=17$, and height $3$. Let $E$ be the intersection of $AC$ and $BD$. Circles $\Omega$ and $\omega$ are circumscribed about triangles $ABE$ and $CDE$. Compute the sum of the radii of $\Omega$ and $\omega$.

Let the medians of the triangle $ABC$ intersect at point $M$. A line $t$ through $M$ intersects the circumcircle of $ABC$ at $X$ and $Y$ so that $A$ and $C$ lie on the same side of $t$. Prove that $BX\cdot BY=AX\cdot AY+CX\cdot CY$.

In a trapezoid $ABCD$ with $AB\parallel CD$, the diagonals $AC$ and $BD$ meet at $O$. Let $M$ and $N$ be the centers of the regular hexagons constructed on the sides $AB$ and $CD$ in the exterior of the trapezoid. Prove that $M,O$ and $N$ are collinear.

A trapezoid with bases $AD$ and $BC$ is circumscribed about a circle, $E$ is the intersection point of the diagonals. Prove that $\angle AED$ is not acute.

Circles with centers $A ,~ B,$ and $C$ each have radius $r$, where $1 < r < 2$. The distance between each pair of centers is $2$. If $B'$ is the point of intersection of circle $A$ and circle $C$ which is outside circle $B$, and if $C'$ is the point of intersection of circle $A$ and circle $B$ which is outside circle $C$, then length $B'C'$ equals $\textbf{(A) }3r-2\qquad\textbf{(B) }r^2\qquad\textbf{(C) }r+\sqrt{3(r-1)}\qquad$ $\textbf{(D) }1+\sqrt{3(r^2-1)}\qquad\textbf{(E) }\text{none of these}$ [asy] //Holy crap, CSE5 is freaking amazing! import cse5; pathpen=black; pointpen=black; dotfactor=3; size(200); pair A=(1,2),B=(2,0),C=(0,0); D(CR(A,1.5)); D(CR(B,1.5)); D(CR(C,1.5)); D(MP("$A$",A)); D(MP("$B$",B)); D(MP("$C$",C)); pair[] BB,CC; CC=IPs(CR(A,1.5),CR(B,1.5)); BB=IPs(CR(A,1.5),CR(C,1.5)); D(BB[0]--CC[1]); MP("$B'$",BB[0],NW);MP("$C'$",CC[1],NE); //Credit to TheMaskedMagician for the diagram [/asy]

A hexagon that is inscribed in a circle has side lengths $22$, $22$, $20$, $22$, $22$, and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$, where $p$ and $q$ are positive integers. Find $p+q$.

Two circles with centers $O_{1}$ and $O_{2}$ meet at two points $A$ and $B$ such that the centers of the circles are on opposite sides of the line $AB$. The lines $BO_{1}$ and $BO_{2}$ meet their respective circles again at $B_{1}$ and $B_{2}$. Let $M$ be the midpoint of $B_{1}B_{2}$. Let $M_{1}$, $M_{2}$ be points on the circles of centers $O_{1}$ and $O_{2}$ respectively, such that $\angle AO_{1}M_{1}= \angle AO_{2}M_{2}$, and $B_{1}$ lies on the minor arc $AM_{1}$ while $B$ lies on the minor arc $AM_{2}$. Show that $\angle MM_{1}B = \angle MM_{2}B$. [i]Ciprus[/i]

Let $ABCD$ be a convex quadrilateral with $AB$ parallel to $CD$. Let $P$ and $Q$ be the midpoints of $AC$ and $BD$, respectively. Prove that if $\angle ABP=\angle CBD$, then $\angle BCQ=\angle ACD$.

Diagonals of trapezium $ABCD$ are mutually perpendicular and the midline of the trapezium is $5$. Find the length of the segment that connects the midpoints of the bases of the trapezium.

Let side $BC$ of $\bigtriangleup ABC$ be the diameter of a semicircle which cuts $AB$ and $AC$ at $D$ and $E$ respectively. $F$ and $G$ are the feet of the perpendiculars from $D$ and $E$ to $BC$ respectively. $DG$ and $EF$ intersect at $M$. Prove that $AM \perp BC$.

The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with $ 9$ trapezoids, let $ x$ be the angle measure in degrees of the larger interior angle of the trapezoid. What is $ x$? [asy]unitsize(4mm); defaultpen(linewidth(.8pt)); int i; real r=5, R=6; path t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0); for(i=0; i<9; ++i) { draw(rotate(20*i)*t); } draw((-r,0)--(R+1,0)); draw((-R,0)--(-R-1,0));[/asy]$ \textbf{(A)}\ 100 \qquad \textbf{(B)}\ 102 \qquad \textbf{(C)}\ 104 \qquad \textbf{(D)}\ 106 \qquad \textbf{(E)}\ 108$

In the trapezoid $ABCD$ , the base $AB$ is smaller than the $CD$ base. The point $K$ is chosen such that $AK$ is parallel to BC and $BK$ is parallel to $AD$. The points $P$ and $Q$ are chosen on the $AK$ and $BK$ rays respectively, such that $\angle ADP = \angle BCK$ and $\angle BCQ = \angle ADK$. (a) Show that the lines $AD, BC$ and $PQ$ go through the same point. (b) Assuming that the circumscribed circumferences of the $APD$ and $BCQ$ triangles intersect at two points, show that one of those points belongs to the line $PQ$.

Given quadrilateral $ ABCD$. The circumcircle of $ ABC$ is tangent to side $ CD$, and the circumcircle of $ ACD$ is tangent to side $ AB$. Prove that the length of diagonal $ AC$ is less than the distance between the midpoints of $ AB$ and $ CD$.