Let $ABC$ be an acute,non-isosceles triangle with $AB<AC<BC$.Let $D,E,Z$ be the midpoints of $BC,AC,AB$ respectively and segments $BK,CL$ are altitudes.In the extension of $DZ$ we take a point $M$ such that the parallel from $M$ to $KL$ crosses the extensions of $CA,BA,DE$ at $S,T,N$ respectively (we extend $CA$ to $A$-side and $BA$ to $A$-side and $DE$ to $E$-side).If the circumcirle $(c_{1})$ of $\triangle{MBD}$ crosses the line $DN$ at $R$ and the circumcirle $(c_{2})$ of $\triangle{NCD}$ crosses the line $DM$ at $P$ prove that $ST\parallel PR$.

$ABCDE$ is convex pentagon. $m(\widehat{B})=m(\widehat{D})=90^\circ$, $m(\widehat{C})=120^\circ$, $|AB|=2$, $|BC|=|CD|=\sqrt3$, and $|ED|=1$. $|AE|=?$ $ \textbf{(A)}\ \frac{3\sqrt3}{2} \qquad\textbf{(B)}\ \frac{2\sqrt3}{3} \qquad\textbf{(C)}\ \frac{3}{2} \qquad\textbf{(D)}\ \sqrt3 - 1 \qquad\textbf{(E)}\ \sqrt3 $

Let $ABCD$ be a trapezium with $AD // BC$. Suppose $K$ and $L$ are, respectively, points on the sides $AB$ and $CD$ such that $\angle BAL = \angle CDK$. Prove that $\angle BLA = \angle CKD$.

Given that points $ D,E$ lie on the sidelines $ AB,BC$ of triangle $ ABC$, respectively, point $ P$ is in interior of triangle $ ABC$ such that $ PE \equal{} PC$ and $ \bigtriangleup DEP\sim \bigtriangleup PCA.$ Prove that $ BP$ is tangent of the circumcircle of triangle $ PAD.$

Given is trapezoid $ABCD$, $M$ and $N$ being the midpoints of the bases of $AD$ and $BC$, respectively. a) Prove that the trapezoid is isosceles if it is known that the intersection point of perpendicular bisectors of the lateral sides belongs to the segment $MN$. b) Does the statement of point a) remain true if it is only known that the intersection point of perpendicular bisectors of the lateral sides belongs to the line $MN$?

Let $ABCD$ be a trapezoid with bases $AB$ and $CD$. Points $P$ and $Q$ lie on diagonals $AC$ and $BD$, respectively and $\angle APD = \angle BQC$. Prove that $\angle AQD = \angle BPC$.

Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE \equal{} \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$. [i]Proposed by Charles Leytem, Luxembourg[/i]

$n$ is a positive integer. An equilateral triangle of side length $3n$ is split into $9n^2$ unit equilateral triangles, each colored one of red, yellow, blue, such that each color appears $3n^2$ times. We call a trapezoid formed by three unit equilateral triangles as a "standard trapezoid". If a "standard trapezoid" contains all three colors, we call it a "colorful trapezoid". Find the maximum possible number of "colorful trapezoids".

Trapezoid $ ABCD$ has bases $ \overline{AB}$ and $ \overline{CD}$ and diagonals intersecting at $ K$. Suppose that $ AB\equal{}9$, $ DC\equal{}12$, and the area of $ \triangle AKD$ is $ 24$. What is the area of trapezoid $ ABCD$? $ \textbf{(A)}\ 92 \qquad \textbf{(B)}\ 94 \qquad \textbf{(C)}\ 96 \qquad \textbf{(D)}\ 98 \qquad \textbf{(E)}\ 100$

In a convex quadrilateral $ ABCD, $ let $ A’,B’ $ be the orthogonal projections to $ CD $ of $ A, $ respectively, $ B. $ [b]a)[/b] Assuming that $ BB’\le AA’ $ and that the perimeter of $ ABCD $ is $ (AB+CD)\cdot BB’, $ is $ ABCD $ necessarily a trapezoid? [b]b)[/b] The same question with the addition that $ \angle BAD $ is obtuse.

As shown in the figure below, $ABCD$ is a trapezoid, $AB \parallel CD$. The sides $DA$, $AB$, $BC$ are tangent to $\odot O_1$ and $AB$ touches $\odot O_1$ at $P$. The sides $BC$, $CD$, $DA$ are tangent to $\odot O_2$, and $CD$ touches $\odot O_2$ at $Q$. Prove that the lines $AC$, $BD$, $PQ$ meet at the same point. [asy] size(200); defaultpen(linewidth(0.8)+fontsize(10pt)); pair A=origin,B=(1,-7),C=(30,-15),D=(26,6); pair bisA=bisectorpoint(B,A,D),bisB=bisectorpoint(A,B,C),bisC=bisectorpoint(B,C,D),bisD=bisectorpoint(C,D,A); path bA=A--(bisA+100*(bisA-A)),bB=B--(bisB+100*(bisB-B)),bC=C--(bisC+100*(bisC-C)),bD=D--(bisD+100*(bisD-D)); pair O1=intersectionpoint(bA,bB),O2=intersectionpoint(bC,bD); dot(O1^^O2,linewidth(2)); pair h1=foot(O1,A,B),h2=foot(O2,C,D); real r1=abs(O1-h1),r2=abs(O2-h2); draw(circle(O1,r1)^^circle(O2,r2)); draw(A--B--C--D--cycle); draw(A--C^^B--D^^h1--h2); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,dir(350)); label("$D$",D,dir(350)); label("$P$",h1,dir(200)); label("$Q$",h2,dir(350)); label("$O_1$",O1,dir(150)); label("$O_2$",O2,dir(300)); [/asy]

In a trapezoid $ABCD$ with bases $AD$ and $BC$, the bisector of the angle $\angle DAB$ intersects the bisectors of the angles $\angle ABC$ and $\angle CDA$ at the points $P$ and $S$, respectively, and the bisector of the angle $\angle BCD$ intersects the bisectors of the angles $\angle ABC$ and $\angle CDA$ at the points $Q$ and $R$, respectively. Prove that if $PS\parallel RQ$, then $AB = CD$.

Given $n>4$, prove that every cyclic quadrilateral can be dissected into $n$ cyclic quadrilaterals.

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 $MN$.

Five points $A$, $B$, $C$, $D$ and $E$ lie on a circle $\tau$ clockwise in that order such that $AB \parallel CE$ and $\angle ABC > 90^{\circ}$. Let $k$ be a circle tangent to $AD$, $CE$ and $\tau$ such that $k$ and $\tau$ touch on the arc $\widehat{DE}$ not containing $A$, $B$ and $C$. Let $F \neq A$ be the intersection of $\tau$ and the tangent line to $k$ passing through $A$ different from $AD$. Prove that there exists a circle tangent to $BD$, $BF$, $CE$ and $\tau$.