Consider the areas of the four triangles obtained by drawing the diagonals $AC$ and $BD$ of a trapezium $ABCD$. The product of these areas, taken two at time, are computed. If among the six products so obtained, two products are 1296 and 576, determine the square root of the maximum possible area of the trapezium to the nearest integer.

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)

Let $ABCD$ be a trapezoid with base sides $AB$ and $CD$ and let $AB=a$, $BC=b$, $CD=c$, $DA=d$, $AC=m$ and $BD=n$. We know that $m^2+n^2=(a+c)^2$ $a)$ Prove that lines $AC$ and $BD$ are perpendicular $b)$ Prove that $ac<bd$

What is the largest possible area of an isosceles trapezoid in which the largest side is $13$ and the perimeter is $28$? $ \textbf{(A)}\ 13 \qquad\textbf{(B)}\ 24 \qquad\textbf{(C)}\ 27 \qquad\textbf{(D)}\ 28 \qquad\textbf{(E)}\ 30 $

In trapezoid $ABCD$, $AD$ is parallel to $BC$. Knowing that $AB=AD+BC$, prove that the bisector of $\angle A$ also bisects $CD$.

In trapezoid $ABCD$, $BC \parallel AD$, $AB = 13$, $BC = 15$, $CD = 14$, and $DA = 30$. Find the area of $ABCD$.

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 an isosceles trapezoid with bases $AD> BC$. Let $X$ be the intersection of the bisectors of $\angle BAC$ and $BC$. Let $E$ be the intersection of$ DB$ with the parallel to the bisector of $\angle CBD$ through $X$ and let $F$ be the intersection of $DC$ with the parallel to the bisector of $\angle DCB$ through $X$. Show that quadrilateral $AEFD$ is cyclic.

Let $ABCD$ be an isosceles trapezoid such that $AB \parallel CD$. Suppose that there exists a point $P$ in $ABCD$ such that $\angle APB > \angle ADC$ and $\angle DPC > \angle ABC$. Prove that $$AB+CD>DA+BC.$$

Let $ABCD$ be a trapezoid with bases $AB,CD$ such that $CD=k \cdot AB$ ($0<k<1$). Point $P$ is such that $\angle PAB=\angle CAD$ and $\angle PBA=\angle DBC$. Prove that $PA+PB \leq \dfrac{1}{\sqrt{1-k^2}} \cdot AB$.

Points $M$ and $N$ are the midpoints of sides $PA$ and $PB$ of $\triangle PAB$. As $P$ moves along a line that is parallel to side $AB$, how many of the four quantities listed below change? $\mathrm{(A)}\ \text{the length of the segment} MN$ $\mathrm{(B)}\ \text{the perimeter of }\triangle PAB$ $\mathrm{(C)}\ \text{ the area of }\triangle PAB$ $\mathrm{(D)}\ \text{ the area of trapezoid} ABNM$ [asy] draw((2,0)--(8,0)--(6,4)--cycle); draw((4,2)--(7,2)); draw((1,4)--(9,4),Arrows); label("$A$",(2,0),SW); label("$B$",(8,0),SE); label("$M$",(4,2),W); label("$N$",(7,2),E); label("$P$",(6,4),N);[/asy] $\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 1 \qquad\mathrm{(C)}\ 2 \qquad\mathrm{(D)}\ 3 \qquad\mathrm{(E)}\ 4$

An isosceles trapezoid $ABCD$ $(AB \parallel CD)$ is given. Points $E$ and $F$ lie on the sides $BC$ and $AD$, and the points $M$ and $N$ lie on the segment $EF$ such that $DF = BE$ and $FM = NE$. Let $K$ and $L$ be the foot of perpendicular lines from $M$ and $N$ to $AB$ and $CD$, respectively. Prove that $EKFL$ is a parallelogram. [i]Proposed by Mahdi Etesamifard[/i]

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} $

Triangle $ABC$ is inscribed in circle $O$. Tangent $PD$ is drawn from $A$, $D$ is on ray $BC$, $P$ is on ray $DA$. Line $PU$ ($U \in BD$) intersects circle $O$ at $Q$, $T$, and intersect $AB$ and $AC$ at $R$ and $S$ respectively. Prove that if $QR=ST$, then $PQ=UT$.

Let a triangle $ABC$ inscribed in circle $c(O,R)$ and $D$ an arbitrary point on $BC$(different from the midpoint).The circumscribed circle of $BOD$,which is $(c_1)$, meets $c(O,R)$ at $K$ and $AB$ at $Z$.The circumscribed circle of $COD$ $(c_2)$,meets $c(O,R)$ at $M$ and $AC$ at $E$.Finally, the circumscribed circle of $AEZ$ $(c_3)$,meets $c(O,R)$ at $N$.Prove that $\triangle{ABC}=\triangle{KMN}.$

Let $ABC$ be a triangle with circumcentre $O$. The points $D,E,F$ lie in the interiors of the sides $BC,CA,AB$ respectively, such that $DE$ is perpendicular to $CO$ and $DF$ is perpendicular to $BO$. (By interior we mean, for example, that the point $D$ lies on the line $BC$ and $D$ is between $B$ and $C$ on that line.) Let $K$ be the circumcentre of triangle $AFE$. Prove that the lines $DK$ and $BC$ are perpendicular. [i]Netherlands (Merlijn Staps)[/i]

On the trapezoid $ABCD$ , side $DA$ is perpendicular to the bases $AB$ and $CD$ . The base $AB$ measures $45$, the base $CD$ measures $20$ and the $BC$ side measures $65$. Let $P$ on the $BC$ side such that $BP$ measures $45$ and $M$ is the midpoint of $DA$. Calculate the measure of the $PM$ segment.

Let $O$ be the circumcenter of the triangle $ABC$. The segment $XY$ is the diameter of the circumcircle perpendicular to $BC$ and it meets $BC$ at $M$. The point $X$ is closer to $M$ than $Y$ and $Z$ is the point on $MY$ such that $MZ = MX$. The point $W$ is the midpoint of $AZ$. a) Show that $W$ lies on the circle through the midpoints of the sides of $ABC$; b) Show that $MW$ is perpendicular to $AY$.

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) $QR \perp AD$.

Let $ABCD$ be a quadrilateral with $\measuredangle DAB=\measuredangle ABC=90^{\circ}$. Denote by $M$ the midpoint of the side $AB$, and assume that $\measuredangle CMD=90^{\circ}$. Let $K$ be the foot of the perpendicular from the point $M$ to the line $CD$. The line $AK$ meets $BD$ at $P$, and the line $BK$ meets $AC$ at $Q$. Show that $\angle{AKB}=90^{\circ}$ and $\frac{KP}{PA}+\frac{KQ}{QB}=1$. [color=red][Moderator edit: The proposed solution can be found at http://erdos.fciencias.unam.mx/mexproblem3.pdf .][/color]

Let $\omega$ be incircle of $ABC$. $P$ and $Q$ are on $AB$ and $AC$, such that $PQ$ is parallel to $BC$ and is tangent to $\omega$. $AB,AC$ touch $\omega$ at $F,E$. Prove that if $M$ is midpoint of $PQ$, and $T$ is intersection point of $EF$ and $BC$, then $TM$ is tangent to $\omega$. [i]By Ali Khezeli[/i]

$ABCD$ is a trapezoid with $BC\parallel AD$ and $BC = 2AD$. Point $M$ is chosen on the side $CD$ such that $AB = AM$. Prove that $BM \perp CD$. [i]Proposed by Bogdan Rublov[/i]

Let $ABCD$ a trapezium with $AB$ parallel to $CD, \Omega$ the circumcircle of $ABCD$ and $A_1,B_1$ points on segments $AC$ and $BC$ respectively, such that $DA_1B_1C$ is a cyclic cuadrilateral. Let $A_2$ and $B_2$ the symmetric points of $A_1$ and $B_1$ with respect of the midpoint of $AC$ and $BC$, respectively. Prove that points $A, B, A_2, B_2$ are concyclic.

Suppose that $ M$ is an arbitrary point on side $ BC$ of triangle $ ABC$. $ B_1,C_1$ are points on $ AB,AC$ such that $ MB = MB_1$ and $ MC = MC_1$. Suppose that $ H,I$ are orthocenter of triangle $ ABC$ and incenter of triangle $ MB_1C_1$. Prove that $ A,B_1,H,I,C_1$ lie on a circle.

9.6, 10.6 Construct for each vertex of the trapezium a symmetric point wrt to the diagonal, which doesn't contain this vertex. Prove that if four new points form a quadrilateral then it is a trapezium. ([i]L. Emel'yanov[/i])