Let $ABC$, $AC \not= BC$, be an acute triangle with orthocenter $H$ and incenter $I$. The lines $CH$ and $CI$ meet the circumcircle of $\bigtriangleup ABC$ at points $D$ and $L$, respectively. Prove that $\angle CIH = 90^{\circ}$ if and only if $\angle IDL = 90^{\circ}$

Let $ABCD$ be a quadrilateral and let $O$ be the point of intersection of diagonals $AC$ and $BD$. Knowing that the area of triangle $AOB$ is equal to $ 1$, the area of triangle $BOC$ is equal to $2$, and the area of triangle $COD$ is equal to $4$, calculate the area of triangle $AOD$ and prove that $ABCD$ is a trapezoid.

Nine lines are given such that every one of them intersects given square $ABCD$ on two trapezoids, which area ratio is $2 : 3$. Prove that at least $3$ of those $9$ lines pass through the same point

Let $ABCD$ be the trapezium with bases $AB$ and $CD$, such that $\sphericalangle ABC = 90^{\circ}$. The bisector of angle $BAD$ intersects the segment $BC$ in the point $P$. Show that if $\sphericalangle APD = 45^{\circ}$, then area of quadrilateral $APCD$ is equal to the area of the triangle $ABP$.

Consider a trapezoid $ABCD$ with bases $AB$ and $CD$ satisfying $| AB | > | CD |$. Let $M$ be the midpoint of $AB$. Let the point $P$ lie inside $ABCD$ such that $| AD | = | PC |$ and $| BC | = | PD |$. Prove that if $| \angle CMD | = 90^o$, then the quadrilaterals $AMPD$ and $BMPC$ have the same area.

The line joining the midpoints of the diagonals of a trapezoid has length $3$. If the longer base is $97$, then the shorter base is: $ \textbf{(A)}\ 94 \qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 89 $

A trapezoid $ABCD$ ($AB ///CD$) has the property that there are points $P$ and $Q$ on sides $AD$ and $BC$ respectively such that $\angle APB = \angle CPD$ and $\angle AQB = \angle CQD$. Show that the points $P$ and $Q$ are equidistant from the intersection point of the diagonals of the trapezoid. (M. Smurov)

Some cells of a checkered plane are marked so that figure $A$ formed by marked cells satisfies the following condition:$1)$ any cell of the figure $A$ has exactly two adjacent cells of $A$; and $2)$ the figure $A$ can be divided into isosceles trapezoids of area $2$ with vertices at the grid nodes (and acute angles of trapezoids are equal to $45$) . Prove that the number of marked cells is divisible by $8$.

Let $ K,L,M$ and $ N$ be the midpoints of sides $ AB,$ $ BC,$ $ CD$ and $ AD$ of the convex quadrangle $ ABCD.$ Is it possible that points $ A,B,L,M,D$ lie on the same circle and points $ K,B,C,D,N$ lie on the same circle?

Let $ABCD$ be a trapezoid such that $|AB|=9$, $|CD|=5$ and $BC\parallel AD$. Let the internal angle bisector of angle $D$ meet the internal angle bisectors of angles $A$ and $C$ at $M$ and $N$, respectively. Let the internal angle bisector of angle $B$ meet the internal angle bisectors of angles $A$ and $C$ at $L$ and $K$, respectively. If $K$ is on $[AD]$ and $\dfrac{|LM|}{|KN|} = \dfrac 37$, what is $\dfrac{|MN|}{|KL|}$? $ \textbf{(A)}\ \dfrac{62}{63} \qquad\textbf{(B)}\ \dfrac{27}{35} \qquad\textbf{(C)}\ \dfrac{2}{3} \qquad\textbf{(D)}\ \dfrac{5}{21} \qquad\textbf{(E)}\ \dfrac{24}{63} $

A circle $k$ and a point $A$ outside it are given in the plane. Prove that all trapezoids, whose non-parallel sides meet at $A$, have the same intersection of diagonals.

The non-parallel sides of a trapezium serve as the diameters of two circles. Prove that all four tangents to the circles drawn from the point of intersection of the diagonals are equal (if this point lies outside the circles). (S Markelov)

Let $ABC$ be a triangle. Prove that there is a line $\ell$ (in the plane of triangle $ABC$) such that the intersection of the interior of triangle $ABC$ and the interior of its reflection $A'B'C'$ in $\ell$ has area more than $\frac23$ the area of triangle $ABC$.

A quadrilateral $ABCD$ is inscribed in a circle $\omega$. The tangent to $\omega$ at $A$ intersects the ray $CB$ at $K$, and the tangent to $\omega$ at $B$ intersects the ray $DA$ at $M$. Prove that if $AM=AD$ and $BK=BC$, then $ABCD$ is a trapezoid.

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 $ST\parallel PR$.

Let $ABCD$ be an isosceles trapezoid with bases $AB=5$ and $CD=7$ and legs $BC=AD=2 \sqrt{10}.$ A circle $\omega$ with center $O$ passes through $A,B,C,$ and $D.$ Let $M$ be the midpoint of segment $CD,$ and ray $AM$ meet $\omega$ again at $E.$ Let $N$ be the midpoint of $BE$ and $P$ be the intersection of $BE$ with $CD.$ Let $Q$ be the intersection of ray $ON$ with ray $DC.$ There is a point $R$ on the circumcircle of $PNQ$ such that $\angle PRC = 45^\circ.$ The length of $DR$ can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. What is $m+n$? [i]Author: Ray Li[/i]

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.

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 $KA=KT$.

$ ABC$ is a triangle with $ \angle BAC \equal{} 10{}^\circ$, $ \angle ABC \equal{} 150{}^\circ$. Let $ X$ be a point on $ \left[AC\right]$ such that $ \left|AX\right| \equal{} \left|BC\right|$. Find $ \angle BXC$. $\textbf{(A)}\ 15^\circ \qquad\textbf{(B)}\ 20^\circ \qquad\textbf{(C)}\ 25^\circ \qquad\textbf{(D)}\ 30^\circ \qquad\textbf{(E)}\ 35^\circ$

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]

Triangle $ABC$ is equilateral with $AB=1$. Points $E$ and $G$ are on $\overline{AC}$ and points $D$ and $F$ are on $\overline{AB}$ such that both $\overline{DE}$ and $\overline{FG}$ are parallel to $\overline{BC}$. Furthermore, triangle $ADE$ and trapezoids $DFGE$ and $FBCG$ all have the same perimeter. What is $DE+FG$? [asy] size(180); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); real s=1/2,m=5/6,l=1; pair A=origin,B=(l,0),C=rotate(60)*l,D=(s,0),E=rotate(60)*s,F=m,G=rotate(60)*m; draw(A--B--C--cycle^^D--E^^F--G); dot(A^^B^^C^^D^^E^^F^^G); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,S); label("$E$",E,NW); label("$F$",F,S); label("$G$",G,NW); [/asy] $\textbf{(A) }1\qquad \textbf{(B) }\dfrac{3}{2}\qquad \textbf{(C) }\dfrac{21}{13}\qquad \textbf{(D) }\dfrac{13}{8}\qquad \textbf{(E) }\dfrac{5}{3}\qquad$

Let $ABCD$ be a convex quadrilateral. The perpendicular bisectors of the sides $[AD]$ and $[BC]$ intersect at a point $P$ inside the quadrilateral and the perpendicular bisectors of the sides $[AB]$ and $[CD]$ also intersect at a point $Q$ inside the quadrilateral. Show that, if $\angle APD = \angle BPC$ then $\angle AQB = \angle CQD$

Let $ABC$ and $PQR$ be two triangles such that [list] [b](a)[/b] $P$ is the mid-point of $BC$ and $A$ is the midpoint of $QR$. [b](b)[/b] $QR$ bisects $\angle BAC$ and $BC$ bisects $\angle QPR$ [/list] Prove that $AB+AC=PQ+PR$.

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 non-isosceles trapezoid. Define a point $A1$ as intersection of circumcircle of triangle $BCD$ and line $AC$. (Choose $A_1$ distinct from $C$). Points $B_1, C_1, D_1$ are defined in similar way. Prove that $A_1B_1C_1D_1$ is a trapezoid as well.