$ABC$ is an acute triangle with orthocenter $H$. Points $D$ and $E$ lie on segment $BC$. Circumcircle of $\triangle BHC$ instersects with segments $AD$,$AE$ at $P$ and $Q$, respectively. Prove that if $BD^2+CD^2=2DP\cdot DA$ and $BE^2+CE^2=2EQ\cdot EA$, then $BP=CQ$.

Let $\omega$ be the circumcircle of a triangle $ABC$. Denote by $M$ and $N$ the midpoints of the sides $AB$ and $AC$, respectively, and denote by $T$ the midpoint of the arc $BC$ of $\omega$ not containing $A$. The circumcircles of the triangles $AMT$ and $ANT$ intersect the perpendicular bisectors of $AC$ and $AB$ at points $X$ and $Y$, respectively; assume that $X$ and $Y$ lie inside the triangle $ABC$. The lines $MN$ and $XY$ intersect at $K$. Prove that $KA=KT$.

The trapezoid $ABCD$ of bases $AB$ and $CD$, has $\angle A = 90^o, AB = 6, CD = 3$ and $AD = 4$. Let $E, G, H$ be the circumcenters of triangles $ABC, ACD, ABD$, respectively. Find the area of the triangle $EGH$.

We consider a right trapezoid $ABCD$, in which $AB||CD,AB>CD,AD\perp AB$ and $AD>CD$. The diagonals $AC$ and $BD$ intersect at $O$. The parallel through $O$ to $AB$ intersects $AD$ in $E$ and $BE$ intersects $CD$ in $F$. Prove that $CE\perp AF$ if and only if $AB\cdot CD=AD^2-CD^2$ .

Let $ABCD$ be a trapezoid of parallel bases $ BC$ and $AD$. If $\angle CAD = 2\angle CAB, BC = CD$ and $AC = AD$, determine all the possible values of the measure of the angle $\angle CAB$.

Let $A'$ be a point on one of the sides of the trapezoid $ABCD$ such that line $AA'$ divides the area of the trapezoid in half. Points $B'$, $C'$, $D'$ are defined similarly. Prove that the intersection points of the diagonals of quadrilaterals $ABCD$ and $A'B'C'D'$ are symmetrical wrt the midpoint of midline of trapezoid $ABCD$.

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 $MN$.

Find the side sizes of an isosceles trapezoid that has longest side $13$ cm, perimeter $28$ cm and area $27$ cm$^2$. Is there such a trapezoid, if we we ask for area $27.001$ cm$^2$ ?

Let $ ABC$ be an acute triangle, $ CC_1$ its bisector, $ O$ its circumcenter. The perpendicular from $ C$ to $ AB$ meets line $ OC_1$ in a point lying on the circumcircle of $ AOB$. Determine angle $ C$.

Given two circles with the radiuses $R$ and $r$, touching each other from the outer side. Consider all the trapezoids, such that its lateral sides touch both circles, and its bases touch different circles. Find the shortest possible lateral side.

A trapezoid is given in which one base is twice as large as the other. Use one ruler (no divisions) to draw the midline of this trapezoid.

Figure $ABCD$ is a trapezoid with $AB || DC, AB = 5, BC = 3 \sqrt 2, \measuredangle BCD = 45^\circ$, and $\measuredangle CDA = 60^\circ$. The length of $DC$ is $\textbf{(A) }7 + \frac{2}{3} \sqrt{3}\qquad \textbf{(B) }8\qquad \textbf{(C) }9 \frac{1}{2}\qquad \textbf{(D) }8 + \sqrt 3\qquad \textbf{(E) }8 + 3 \sqrt 3$

A square $ ABCD$ with sides of length 1 is divided into two congruent trapezoids and a pentagon, which have equal areas, by joining the center of the square with points $ E,F,G$ where $ E$ is the midpoint of $ BC$, $ F,G$ are on $ AB$ and $ CD$, respectively, and they're positioned that $ AF < FB, DG < GC$ and $ F$ is the directly opposite of $ G$. If $ FB \equal{} x$, the length of the longer parallel side of each trapezoid, find the value of $ x$. [asy]unitsize(2.5cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair[] dotted={(0,0),(0,1),(1,1),(1,0),(1/6,0),(1/6,1),(1/2,1/2),(1,1/2)}; draw(unitsquare); draw((1/6,0)--(1/2,1/2)--(1/6,1)); draw((1/2,1/2)--(1,1/2)); dot(dotted); label("$x$",midpoint((1/6,1)--(1,1)),N);[/asy]$ \displaystyle \textbf{(A)}\ \frac {3}{5} \qquad \textbf{(B)}\ \frac {2}{3} \qquad \textbf{(C)}\ \frac {3}{4} \qquad \textbf{(D)}\ \frac {5}{6} \qquad \textbf{(E)}\ \frac {7}{8}$

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]

Two planes, $P$ and $Q$, intersect along the line $p$. The point $A$ is given in the plane $P$, and the point $C$ in the plane $Q$; neither of these points lies on the straight line $p$. Construct an isosceles trapezoid $ABCD$ (with $AB \parallel CD$) in which a circle can be inscribed, and with vertices $B$ and $D$ lying in planes $P$ and $Q$ respectively.

A non-isosceles trapezoid $ABCD$ ($AB // CD$) is given. An arbitrary circle passing through points $A$ and $B$ intersects the sides of the trapezoid at points $P$ and $Q$, and the intersect the diagonals at points $M$ and $N$. Prove that the lines $PQ, MN$ and $CD$ are concurrent.

A circle center $O$ is inscribed in the quadrilateral $ABCD$. $AB$ is parallel to and longer than $CD$ and has midpoint $M$. The line $OM$ meets $CD$ at $F$. $CD$ touches the circle at $E$. Show that $DE = CF$ iff $AB = 2CD$.

A circle $k$ is given with a non-diameter chord $AC$. On the tangent at point $A$ select point $X \ne A$ and mark $D$ the intersection of the circle $k$ with the interior of the line $XC$ (if any). Let $B$ a point in circle $k$ such that quadrilateral $ABCD$ is a trapezoid . Determine the set of intersections of lines $BC$ and $AD$ belonging to all such trapezoids.

Isosceles trapezoid $ABCD$ has parallel sides $\overline{AD}$ and $\overline{BC},$ with $BC < AD$ and $AB = CD.$ There is a point $P$ in the plane such that $PA=1, PB=2, PC=3,$ and $PD=4.$ What is $\tfrac{BC}{AD}?$ $\textbf{(A) }\frac{1}{4}\qquad\textbf{(B) }\frac{1}{3}\qquad\textbf{(C) }\frac{1}{2}\qquad\textbf{(D) }\frac{2}{3}\qquad\textbf{(E) }\frac{3}{4}$

We are given a trapezoid $ABCD$ with $AB \parallel CD$, $CD=2AB$ and $DB \perp BC$. Let $E$ be the intersection of lines $DA$ and $CB$, and $M$ be the midpoint of $DC$. (a) Prove that $ABMD$ is a rhombus. (b) Prove that triangle $CDE$ is isosceles. (c) If $AM$ and $BD$ meet at $O$, and $OE$ and $AB$ meet at $N,$ prove that the line $DN$ bisects segment $EB$.

Triangles $ABC$ and $ABD$ are isosceles with $AB =AC = BD$, and $BD$ intersects $AC$ at $E$. If $BD$ is perpendicular to $AC$, then $\angle C + \angle D$ is [asy] size(130); defaultpen(linewidth(0.8) + fontsize(11pt)); pair A, B, C, D, E; real angle = 70; B = origin; A = dir(angle); D = dir(90-angle); C = rotate(2*(90-angle), A) * B; draw(A--B--C--cycle); draw(B--D--A); E = extension(B, D, C, A); draw(rightanglemark(B, E, A, 1.5)); label("$A$", A, dir(90)); label("$B$", B, dir(210)); label("$C$", C, dir(330)); label("$D$", D, dir(0)); label("$E$", E, 1.5*dir(340)); [/asy] $\textbf{(A)}\ 115^\circ \qquad \textbf{(B)}\ 120^\circ \qquad \textbf{(C)}\ 130^\circ \qquad \textbf{(D)}\ 135^\circ \qquad \textbf{(E)}\ \text{not uniquely determined}$

Find the area of the $MNRK$ trapezoid with the lateral side $RK = 3$ if the distances from the vertices $M$ and $N$ to the line $RK$ are $5$ and $7$, respectively.

Let $ABCD$ be an isosceles trapezoid with $AD = BC = 255$ and $AB = 128$. Let $M$ be the midpoint of $CD$ and let $N$ be the foot of the perpendicular from $A$ to $CD$. If $\angle MBC = 90^o$, compute $\tan\angle NBM$.

Let $ABCD$ be a trapezium, $AD\parallel BC$, and let $E,F$ be points on the sides$AB$ and $CD$, respectively. The circumcircle of $AEF$ meets $AD$ again at $A_1$, and the circumcircle of $CEF$ meets $BC$ again at $C_1$. Prove that $A_1C_1,BD,EF$ are concurrent.

A circle $ \omega$ with center $ O$ is tangent to the rays of an angle $ BAC$ at $ B$ and $ C$. Point $ Q$ is taken inside the angle $ BAC$. Assume that point $ P$ on the segment $ AQ$ is such that $ AQ\perp OP$. The line $ OP$ intersects the circumcircles $ \omega_{1}$ and $ \omega_{2}$ of triangles $ BPQ$ and $ CPQ$ again at points $ M$ and $ N$. Prove that $ OM \equal{} ON$.