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]

In a trapezoid $ABCD$ with $AB<CD$ and $AB \parallel CD$, the diagonals intersect each other at $E$. Let $F$ be the midpoint of the arc $BC$ (not containing the point $E$) of the circumcircle of the triangle $EBC$. The lines $EF$ and $BC$ intersect at $G$. The circumcircle of the triangle $BFD$ intersects the ray $[DA$ at $H$ such that $A \in [HD]$. The circumcircle of the triangle $AHB$ intersects the lines $AC$ and $BD$ at $M$ and $N$, respectively. $BM$ intersects $GH$ at $P$, $GN$ intersects $AC$ at $Q$. Prove that the points $P, Q, D$ are collinear.

In a trapezoid $ABCD$ such that the internal angles are not equal to $90^{\circ}$, the diagonals $AC$ and $BD$ intersect at the point $E$. Let $P$ and $Q$ be the feet of the altitudes from $A$ and $B$ to the sides $BC$ and $AD$ respectively. Circumscribed circles of the triangles $CEQ$ and $DEP$ intersect at the point $F\neq E$. Prove that the lines $AP$, $BQ$ and $EF$ are either parallel to each other, or they meet at exactly one point.

Let $ABCD$ be a trapezoid ($AD//BC$) with $\angle A=\angle B= 90^o$ and $AD<BC$. Let $E$ be the intersection point of the non parallel sides $AB$ and $CD$, $Z$ be the symmetric point of $A$ wrt line $BC$ and $M$ be the midpoint of $EZ$. If it is given than line $CM$ is perpendicular on line $DZ$, then prove that line $ZC$ is perpendicular on line $EC$.

Let $ABCD$ be the isosceles trapezium with bases $AB$ and $CD$, such that $AC = BC$. The point $M$ is the midpoint of side $AD$. Prove that $\sphericalangle ACM = \sphericalangle CBD$.

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 $O_1 E O_2$.

Let $ ABCD$ be a (not self-intersecting) quadrilateral satisfying $ \measuredangle DAB \equal{} \measuredangle BCD\neq 90^{\circ}$. Let $ X$ and $ Y$ be the orthogonal projections of the point $ D$ on the lines $ AB$ and $ BC$, and let $ Z$ and $ W$ be the orthogonal projections of the point $ B$ on the lines $ CD$ and $ DA$. Establish the following facts: [b]a)[/b] The quadrilateral $ XYZW$ is an isosceles trapezoid such that $ XY\parallel ZW$. [b]b)[/b] Let $ M$ be the midpoint of the segment $ AC$. Then, the lines $ XZ$ and $ YW$ pass through the point $ M$. [b]c)[/b] Let $ N$ be the midpoint of the segment $ BD$, and let $ X^{\prime}$, $ Y^{\prime}$, $ Z^{\prime}$, $ W^{\prime}$ be the midpoints of the segments $ AB$, $ BC$, $ CD$, $ DA$. Then, the point $ M$ lies on the circumcircles of the triangles $ W^{\prime}X^{\prime}N$ and $ Y^{\prime}Z^{\prime}N$. [hide="Notice"][i]Notice.[/i] This problem has been discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=172417 .[/hide]

Let $ ABC$ be a triangle, $ D$ be the midpoint of $ BC$, $ E$ be a point on segment $ AC$ such that $ BE\equal{}2AD$ and $ F$ is the intersection point of $ AD$ with $ BE$. If $ \angle DAC\equal{}60^{\circ}$, find the measure of the angle $ FEA$.

In a convex quadrilateral $ABCD$, the diagonals are perpendicular to each other and they intersect at $E$. Let $P$ be a point on the side $AD$ which is different from $A$ such that $PE=EC.$ The circumcircle of triangle $BCD$ intersects the side $AD$ at $Q$ where $Q$ is also different from $A$. The circle, passing through $A$ and tangent to line $EP$ at $P$, intersects the line segment $AC$ at $R$. If the points $B, R, Q$ are concurrent then show that $\angle BCD=90^{\circ}$.

Is there exist four points on plane, such that distance between any two of them is integer odd number? May be it is geometry or number theory or combinatoric, I don't know, so it here :blush:

Let $ABCS$ be an isosceles trapezoid. Denote $A',B',C',D'$ the incenters of triangles $BCD,CDA,$ $DAB,ABC,$ respectively. Show that $A',B',C',D'$ are vertices of a rectangle.

A circle centered at a point $F$ and a parabola with focus $F$ have two common points. Prove that there exist four points $A, B, C, D$ on the circle such that the lines $AB, BC, CD$ and $DA$ touch the parabola.

Given is an isosceles trapezoid $ABCD$ with $AB \parallel CD$ and $AB> CD$. The projection from $D$ on $ AB$ is $E$. The midpoint of the diagonal $BD$ is $M$. Prove that $EM$ is parallel to $AC$. (Karl Czakler)

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 $BC^2.$

Square $ AIME$ has sides of length $ 10$ units. Isosceles triangle $ GEM$ has base $ EM$, and the area common to triangle $ GEM$ and square $ AIME$ is $ 80$ square units. Find the length of the altitude to $ EM$ in $ \triangle GEM$.

From which two statements about the trapezoid follows the third: 1) the trapezoid is tangential, 2) the trapezoid is right, 3) its area is equal to the product of the bases?

Let $ABCD$ be a square of side length $4$. Points $E$ and $F$ are chosen on sides $BC$ and $DA$, respectively, such that $EF = 5$. Find the sum of the minimum and maximum possible areas of trapezoid $BEDF$. [i]Proposed by Andrew Wu[/i]

Let $ A_{1}A_{2}A_{3}A_{4}A_{5}$ be a convex pentagon, such that \[ [A_{1}A_{2}A_{3}] \equal{} [A_{2}A_{3}A_{4}] \equal{} [A_{3}A_{4}A_{5}] \equal{} [A_{4}A_{5}A_{1}] \equal{} [A_{5}A_{1}A_{2}].\] Prove that there exists a point $ M$ in the plane of the pentagon such that \[ [A_{1}MA_{2}] \equal{} [A_{2}MA_{3}] \equal{} [A_{3}MA_{4}] \equal{} [A_{4}MA_{5}] \equal{} [A_{5}MA_{1}].\] Here $ [XYZ]$ stands for the area of the triangle $ \Delta XYZ$.

An equilateral triangle whose side is $ 2$ is divided into a triangle and a trapezoid by a line drawn parallel to one of its sides. If the area of the trapezoid equals one-half of the area of the original triangle, the length of the median of the trapezoid is: $ \textbf{(A)}\ \frac{\sqrt{6}}{2} \qquad \textbf{(B)}\ \sqrt{2} \qquad \textbf{(C)}\ 2\plus{}\sqrt{2} \qquad \textbf{(D)}\ \frac{2\plus{}\sqrt{2}}{2} \qquad \textbf{(E)}\ \frac{2\sqrt{3}\minus{}\sqrt{6}}{2}$

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.

An isosceles trapezoid $ABCD$ ($AB = CD$) is given. A point $P$ on its circumcircle is such that segments $CP$ and $AD$ meet at point $Q$. Let $L$ be tha midpoint of$ QD$. Prove that the diagonal of the trapezoid is not greater than the sum of distances from the midpoints of the lateral sides to ana arbitrary point of line $PL$.

Consider a regular heptagon ( polygon of $7$ equal sides and angles) $ABCDEFG$ as in the figure below:- $(a).$ Prove $\frac{1}{\sin\frac{\pi}{7}}=\frac{1}{\sin\frac{2\pi}{7}}+\frac{1}{\sin\frac{3\pi}{7}}$ $(b).$ Using $(a)$ or otherwise, show that $\frac{1}{AG}=\frac{1}{AF}+\frac{1}{AE}$ [asy] draw(dir(360/7)..dir(2*360/7),blue); draw(dir(2*360/7)..dir(3*360/7),blue); draw(dir(3*360/7)..dir(4*360/7),blue); draw(dir(4*360/7)..dir(5*360/7),blue); draw(dir(5*360/7)..dir(6*360/7),blue); draw(dir(6*360/7)..dir(7*360/7),blue); draw(dir(7*360/7)..dir(360/7),blue); draw(dir(2*360/7)..dir(4*360/7),blue); draw(dir(4*360/7)..dir(1*360/7),blue); label("$A$",dir(4*360/7),W); label("$B$",dir(5*360/7),S); label("$C$",dir(6*360/7),S); label("$D$",dir(7*360/7),E); label("$E$",dir(1*360/7),E); label("$F$",dir(2*360/7),N); label("$G$",dir(3*360/7),W); [/asy]

Let $MAZN$ be an isosceles trapezium inscribed in a circle $(c)$ with centre $O$. Assume that $MN$ is a diameter of $(c)$ and let $ B$ be the midpoint of $AZ$. Let $(\epsilon)$ be the perpendicular line on $AZ$ passing through $ A$. Let $C$ be a point on $(\epsilon)$, let $E$ be the point of intersection of $CB$ with $(c)$ and assume that $AE$ is perpendicular to $CB$. Let $D$ be the point of intersection of $CZ$ with $(c)$ and let $F$ be the antidiametric point of $D$ on $(c)$. Let $ P$ be the point of intersection of $FE$ and $CZ$. Assume that the tangents of $(c)$ at the points $M$ and $Z$ meet the lines $AZ$ and $PA$ at the points $K$ and $T$ respectively. Prove that $OK$ is perpendicular to $TM$. Theoklitos Parayiou, Cyprus

Let $ ABC$ be a triangle and $ K$ and $ L$ be two points on $ (AB)$, $ (AC)$ such that $ BK \equal{} CL$ and let $ P \equal{} CK\cap BL$. Let the parallel through $ P$ to the interior angle bisector of $ \angle BAC$ intersect $ AC$ in $ M$. Prove that $ CM \equal{} AB$.

A circle $\omega$ is inscribed in a quadrilateral $ABCD$. Let $I$ be the center of $\omega$. Suppose that \[ (AI + DI)^2 + (BI + CI)^2 = (AB + CD)^2. \] Prove that $ABCD$ is an isosceles trapezoid.