Let $ABCD$ be a trapezium such that the sides $AB$ and $CD$ are parallel and the side $AB$ is longer than the side $CD$. Let $M$ and $N$ be on the segments $AB$ and $BC$ respectively, such that each of the segments $CM$ and $AN$ divides the trapezium in two parts of equal area. Prove that the segment $MN$ intersects the segment $BD$ at its midpoint.

A trapezoid $ABCD$ is given with $BC // AD$. Assume that the bisectors of the angles $BAD$ and $CDA$ intersect on the perpendicular bisector of the line segment $BC$. Prove that $|AB|= |CD|$ or $|AB| +|CD| =|AD|$.

An isosceles trapezoid is circumscribed around a circle. The longer base of the trapezoid is $16$, and one of the base angles is $\arcsin(.8)$. Find the area of the trapezoid. $ \textbf{(A)}\ 72\qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 80\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ \text{not uniquely determined} $

In right $ \triangle ABC$ with hypotenuse $ \overline{AB}$, $ AC \equal{} 12$, $ BC \equal{} 35$, and $ \overline{CD}$ is the altitude to $ \overline{AB}$. Let $ \omega$ be the circle having $ \overline{CD}$ as a diameter. Let $ I$ be a point outside $ \triangle ABC$ such that $ \overline{AI}$ and $ \overline{BI}$ are both tangent to circle $ \omega$. The ratio of the perimeter of $ \triangle ABI$ to the length $ AB$ can be expressed in the form $ \displaystyle\frac{m}{n}$, where $ m$ and $ n$ are relatively prime positive integers. Find $ m\plus{}n$.

Let circles $ \Gamma $ and $ \omega $ are circumcircle and incircle of the triangle $ABC$, the incircle touches sides $BC,CA,AB$ at the points $A_1,B_1,C_1$. Let $A_2$ and $B_2$ lies the lines $A_1I$ and $B_1I$ ($A_1$ and $A_2$ lies different sides from $I$, $B_1$ and $B_2$ lies different sides from $I$) such that $IA_2=IB_2=R$. Prove that : (a) $AA_2=BB_2=IO$; (b) The lines $AA_2$ and $BB_2$ intersect on the circle $ \Gamma ;$

In an isosceles trapezoid, the diagonals are perpendicular. Find the distance from the center of the circle described around the trapezoid to the point of intersection of its diagonals, if the lengths of the bases are equal to $a$ and $b$.

Let $ABCD$ be a trapezoid with $AB$ parallel to $CD$ and perpendicular to $BC$. Let $M$ be a point on $BC$ such that $\angle AMB = \angle DMC$. If $AB = 3$, $BC = 24$, and $CD = 4$, what is the value of $AM + MD$?

In an isosceles trapezoid, one of the bases is three times larger than the other. Angle at a greater basis is equal to $45^o$. Show how to cut this trapezium into three parts and make a square with them. Justify your answer.

Quadrilateral $ABCD$ is both cyclic and circumscribed. Its incircle touches its sides $AB$ and $CD$ at points $X$ and $Y$, respectively. The perpendiculars to $AB$ and $CD$ drawn at $A$ and $D$, respectively, meet at point $U$; those drawn at $X$ and $Y$ meet at point $V$, and finally, those drawn at $B$ and $C$ meet at point $W$. Prove that points $U$, $V$ and $W$ are collinear. [i]Proposed by A. Golovanov[/i]

Let $ABCD$ be a trapezoid with bases $AB$ and $CD$. Circle of diameter $BC$ is tangent to line $AD$. Prove, that circle of diameter $AD$ is tangent to line $BC$.

$ABCD$ is a trapezium such that $AB\parallel DC$ and $\frac{AB}{DC}=\alpha >1$. Suppose $P$ and $Q$ are points on $AC$ and $BD$ respectively, such that \[\frac{AP}{AC}=\frac{BQ}{BD}=\frac{\alpha -1}{\alpha+1}\] Prove that $PQCD$ is a parallelogram.

Let $PRUS$ be a trapezium such that $\angle PSR = 2\angle QSU$ and $\angle SPU = 2 \angle UPR$. Let $Q$ and $T$ be on $PR$ and $SU$ respectively such that $SQ$ and $PU$ bisect $\angle PSR$ and $\angle SPU$ respectively. Let $PT$ meet $SQ$ at $E$. The line through $E$ parallel to $SR$ meets $PU$ in $F$ and the line through $E$ parallel to $PU$ meets $SR$ in $G$. Let $FG$ meet $PR$ and $SU$ in $K$ and $L$ respectively. Prove that $KF$ = $FG$ = $GL$.

The isosceles trapezoid $ABCD$ has perpendicular diagonals. The parallel to the bases through the intersection point of the diagonals intersects the non-parallel sides $[BC]$ and $[AD]$ in the points $P$, respectively $R$. The point $Q$ is symmetric of the point $P$ with respect to the midpoint of the segment $[BC]$. Prove that: a) $QR = AD$, b) $QR \perp AD$.

A trapezoid $ABCD$ with bases $BC = a$ and $AD = 2a$ is drawn on the plane. Using only with a ruler, construct a triangle whose area is equal to the area of the trapezoid. With the help of a ruler you can draw straight lines through two known points. (Rozhkova Maria)

A trapezoid $ABCD$ with $BC // AD$ is given. The points $B'$ and $C'$ are symmetrical to $B$ and $C$ with respect to $CD$ and $AB$, respectively. Prove that the midpoint of the segment joining the circumcentres of $ABC'$ and $B'CD$ is equidistant from $A$ and $D$.

The diagram below shows a regular hexagon with an inscribed square where two sides of the square are parallel to two sides of the hexagon. There are positive integers $m$, $n$, and $p$ such that the ratio of the area of the hexagon to the area of the square can be written as $\tfrac{m+\sqrt{n}}{p}$ where $m$ and $p$ are relatively prime. Find $m + n + p$. [asy] import graph; size(4cm); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); draw((0,1)--(1,1)--(1.5,1.87)--(1,2.73)--(0,2.73)--(-0.5,1.87)--cycle); filldraw((1.13,2.5)--(-0.13,2.5)--(-0.13,1.23)--(1.13,1.23)--cycle,grey); draw((0,1)--(1,1)); draw((1,1)--(1.5,1.87)); draw((1.5,1.87)--(1,2.73)); draw((1,2.73)--(0,2.73)); draw((0,2.73)--(-0.5,1.87)); draw((-0.5,1.87)--(0,1)); draw((1.13,2.5)--(-0.13,2.5)); draw((-0.13,2.5)--(-0.13,1.23)); draw((-0.13,1.23)--(1.13,1.23)); draw((1.13,1.23)--(1.13,2.5)); [/asy]

Consider $6$ points located at $P_0=(0,0), P_1=(0,4), P_2=(4,0), P_3=(-2,-2), P_4=(3,3), P_5=(5,5)$. Let $R$ be the region consisting of [i]all[/i] points in the plane whose distance from $P_0$ is smaller than that from any other $P_i$, $i=1,2,3,4,5$. Find the perimeter of the region $R$.

Quadrilateral $ABCD$ has right angles at $B$ and $C$, $\triangle ABC \sim \triangle BCD$, and $AB > BC$. There is a point $E$ in the interior of $ABCD$ such that $\triangle ABC \sim \triangle CEB$ and the area of $\triangle AED$ is $17$ times the area of $\triangle CEB$. What is $\tfrac{AB}{BC}$? $\textbf{(A) \ } 1+\sqrt{2} \qquad \textbf{(B) \ } 2+\sqrt{2}\qquad \textbf{(C) \ } \sqrt{17}\qquad \textbf{(D) \ } 2+\sqrt{5} \qquad \textbf{(E) \ } 1+2\sqrt{3}$

Let $AB$ be a segment of length $26$, and let points $C$ and $D$ be located on $AB$ such that $AC=1$ and $AD=8$. Let $E$ and $F$ be points on one of the semicircles with diameter $AB$ for which $EC$ and $FD$ are perpendicular to $AB$. Find $EF$. $\textbf{(A) }5\qquad\textbf{(B) }5\sqrt2\qquad\textbf{(C) }7\qquad\textbf{(D) }7\sqrt2\qquad\textbf{(E) }12$

On square $ABCD,$ points $E,F,G,$ and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34.$ Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P,$ and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$. [asy] size(200); defaultpen(linewidth(0.8)+fontsize(10.6)); pair A = (0,sqrt(850)); pair B = (0,0); pair C = (sqrt(850),0); pair D = (sqrt(850),sqrt(850)); draw(A--B--C--D--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); pair H = ((2sqrt(850)-sqrt(120))/6,sqrt(850)); pair F = ((2sqrt(850)+sqrt(306)+7)/6,0); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); draw(H--F); pair E = (0,(sqrt(850)-6)/2); pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); draw(E--G); pair P = extension(H,F,E,G); dot("$P$",P,dir(60)); label("$w$", (H+E)/2,fontsize(15)); label("$x$", (E+F)/2,fontsize(15)); label("$y$", (G+F)/2,fontsize(15)); label("$z$", (H+G)/2,fontsize(15)); label("$w:x:y:z=269:275:405:411$",(sqrt(850)/2,-4.5),fontsize(11)); [/asy]

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]

In trapezoid $ABCD,$ leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD},$ and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001},$ find $BC^2.$

The bases of a trapezoid have lengths 10 and 21, and the legs have lengths $\sqrt{34}$ and $3 \sqrt{5}$. What is the area of the trapezoid?

Let a quardilateral $ABCD$ with $AB=AD$ and $\widehat B=\widehat D=90$. At $CD$ we take point $E$ and at $BC$ we take point $Z$ such that $AE\bot DZ$. Prove that $AZ\bot BE$

Let $ABC$ be a nonisosceles triangle. Point $O$ is its circumcenter, and point $K$ is the center of the circumcircle $w$ of triangle $BCO$. The altitude of $ABC$ from $A$ meets $w$ at a point $P$. The line $PK$ intersects the circumcircle of $ABC$ at points $E$ and $F$. Prove that one of the segments $EP$ and $FP$ is equal to the segment $PA$.