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.$

A trapezoid $ABCD$ with bases $AB$ and $CD$ is such that the circumcircle of the triangle $BCD$ intersects the line $AD$ in a point $E$, distinct from $A$ and $D$. Prove that the circumcircle oF the triangle $ABE$ is tangent to the line $BC$.

A checkered polygon $A$ is drawn on the checkered plane. We call a cell of $A$ [i]internal[/i] if all $8$ of its adjacent cells belong to $A$. All other (non-internal) cells of $A$ we call [i]boundary[/i]. It is known that $1)$ each boundary cell has exactly two common sides with no boundary cells; and 2) the union of all boundary cells can be divided into isosceles trapezoid of area $2$ with vertices at the grid nodes (and acute angles of the trapezoids are equal $45^\circ$). Prove that the area of the polygon $A$ is congruent to $1$ modulo $4$.

Let $ABCD$ be an orthodiagonal trapezoid such that $\measuredangle A = 90^{\circ}$ and $AB$ is the larger base. The diagonals intersect at $O$, $\left( OE \right.$ is the bisector of $\measuredangle AOD$, $E \in \left( AD \right)$ and $EF \| AB$, $F \in \left( BC \right)$. Let $P,Q$ the intersections of the segment $EF$ with $AC,BD$. Prove that: (a) $EP=QF$; (b) $EF=AD$. [i]Claudiu-Stefan Popa[/i]

Let $ ABCD$ be an isosceles trapezoid with $ \overline{AD}\parallel{}\overline{BC}$ whose angle at the longer base $ \overline{AD}$ is $ \dfrac{\pi}{3}$. The diagonals have length $ 10\sqrt {21}$, and point $ E$ is at distances $ 10\sqrt {7}$ and $ 30\sqrt {7}$ from vertices $ A$ and $ D$, respectively. Let $ F$ be the foot of the altitude from $ C$ to $ \overline{AD}$. The distance $ EF$ can be expressed in the form $ m\sqrt {n}$, where $ m$ and $ n$ are positive integers and $ n$ is not divisible by the square of any prime. Find $ m \plus{} n$.

In acute triangle $ABC$ the bisector of $\measuredangle A$ meets side $BC$ at $D$. The circle with center $B$ and radius $BD$ intersects side $AB$ at $M$; and the circle with center $C$ and radius $CD$ intersects side $AC$ at $N$. Then it is always true that $ \textbf{(A)}\ \measuredangle CND+\measuredangle BMD-\measuredangle DAC =120^{\circ} \qquad\textbf{(B)}\ AMDN\ \text{is a trapezoid} \qquad\textbf{(C)}\ BC\ \text{is parallel to}\ MN \\ \qquad\textbf{(D)}\ AM-AN=\frac{3(DB-DC)}{2} \qquad\textbf{(E)}\ AB-AC=\frac{3(DB-DC)}{2}$

Let $ABC$ be a fixed triangle. Three similar (by point order) isosceles trapezoids are built on its sides: $ABXY, BCZW, CAUV$, such that the sides of the triangle are bases of the respective trapezoids. The circumcircles of triangles $XZU, YWV$ meet at two points $P, Q$. Prove that the line $PQ$ passes through a fixed point independent of the choice of trapezoids.

Let $ABCD$ be a rectangle. Let $K,L$ be the midpoints of $BC, AD$ respectively. From point $B$ the perpendicular line on $AK$, intersects $AK$ at point $E$ and $CL$ at point $Z$. a) Prove that the quadrilateral $AKZL$ is an isosceles trapezoid b) Prove that $2S_{ABKZ}=S_{ABCD}$ c) If quadrilateral $ABCD$ is a square of side $a$, calculate the area of the isosceles trapezoid $AKZL$ in terms of side $BC=a$

Given an angle with vertex $B$. Construct point $M$ as follows. Let us take an arbitrary isosceles trapezoid whose sides lie on the sides of a given angle. Through two opposite ones draw tangents to the vertices of the circle circumscribed around it. Let $M$ denote the point of intersection of these tangents. What figure do all such points $M$ form?

A trapezoid $ABCD$ is bicentral. The vertex $A$, the incenter $I$, the circumcircle $\omega$ and its center $O$ are given and the trapezoid is erased. Restore it using only a ruler.

let $ABCD$ be a isosceles trapezium having an incircle with $AB$ parallel to $CD$. let $CE$ be the perpendicular from $C$ on $AB$ prove that $ CE^2 = AB. CD $

From the vertices of a regular 27-gon, seven are chosen arbitrarily. Prove that among these seven points there are three points that form an isosceles triangle or four points that form an isosceles trapezoid.

Let $E$ be a point inside the rhombus $ABCD$ such that $|AE|=|EB|$, $m(\widehat{EAB})=12^\circ$, and $m(\widehat{DAE})=72^\circ$. What is $m(\widehat{CDE})$ in degrees? $ \textbf{(A)}\ 64 \qquad\textbf{(B)}\ 66 \qquad\textbf{(C)}\ 68 \qquad\textbf{(D)}\ 70 \qquad\textbf{(E)}\ 72 $

What is the largest possible area of a quadrilateral with sides $1,4,7,8$ ? $ \textbf{(A)}\ 7\sqrt 2 \qquad\textbf{(B)}\ 10\sqrt 3 \qquad\textbf{(C)}\ 18 \qquad\textbf{(D)}\ 12\sqrt 3 \qquad\textbf{(E)}\ 9\sqrt 5 $

In a trapezoid $ABCD$ with $AD \parallel BC , O_{1}, O_{2}, O_{3}, O_{4}$ denote the circles with diameters $AB, BC, CD, DA$, respectively. Show that there exists a circle with center inside the trapezoid which is tangent to all the four circles $O_{1},..., O_{4}$ if and only if $ABCD$ is a parallelogram.

$ABCDE$ 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 $

Quadrilateral $ABCD$ is both cyclic and circumscribed. Its incircle touches its sides $AB$ and $CD$ at points $X$ and $Y$, respectively. The perpendiculars to $AB$ and $CD$ drawn at $A$ and $D$, respectively, meet at point $U$; those drawn at $X$ and $Y$ meet at point $V$, and finally, those drawn at $B$ and $C$ meet at point $W$. Prove that points $U$, $V$ and $W$ are collinear. [i]Proposed by A. Golovanov[/i]

Suppose five of the nine vertices of a regular nine-sided polygon are arbitrarily chosen. Show that one can select four among these five such that they are the vertices of a trapezium.

$\textbf{Bake Sale}$ Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown. $\circ$ Art's cookies are trapezoids: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(5,0)--(5,3)--(2,3)--cycle); draw(rightanglemark((5,3), (5,0), origin)); label("5 in", (2.5,0), S); label("3 in", (5,1.5), E); label("3 in", (3.5,3), N);[/asy] $\circ$ Roger's cookies are rectangles: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(4,0)--(4,2)--(0,2)--cycle); draw(rightanglemark((4,2), (4,0), origin)); draw(rightanglemark((0,2), origin, (4,0))); label("4 in", (2,0), S); label("2 in", (4,1), E);[/asy] $\circ$ Paul's cookies are parallelograms: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(3,0)--(2.5,2)--(-0.5,2)--cycle); draw((2.5,2)--(2.5,0), dashed); draw(rightanglemark((2.5,2),(2.5,0), origin)); label("3 in", (1.5,0), S); label("2 in", (2.5,1), W);[/asy] $\circ$ Trisha's cookies are triangles: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(3,0)--(3,4)--cycle); draw(rightanglemark((3,4),(3,0), origin)); label("3 in", (1.5,0), S); label("4 in", (3,2), E);[/asy] Each friend uses the same amount of dough, and Art makes exactly 12 cookies. How many cookies will be in one batch of Trisha's cookies? $ \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 24$

Rectangle $ ABCD$ and a semicircle with diameter $ AB$ are coplanar and have nonoverlapping interiors. Let $ \mathcal{R}$ denote the region enclosed by the semicircle and the rectangle. Line $ \ell$ meets the semicircle, segment $ AB$, and segment $ CD$ at distinct points $ N$, $ U$, and $ T$, respectively. Line $ \ell$ divides region $ \mathcal{R}$ into two regions with areas in the ratio $ 1: 2$. Suppose that $ AU \equal{} 84$, $ AN \equal{} 126$, and $ UB \equal{} 168$. Then $ DA$ can be represented as $ m\sqrt {n}$, where $ m$ and $ n$ are positive integers and $ n$ is not divisible by the square of any prime. Find $ m \plus{} n$.

Let $AKL$ be a triangle such that $\angle ALK > 90^o +\angle LAK$. Construct an isosceles trapezoid $ABCD$ with $AB \parallel CD$ such that $K$ lies on the side $BC, L$ on the diagonal $AC$ and the lines $AK$ and $BL$ intersect at the circumcenter of the trapezoid.

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 $AB+AC=PQ+PR$.

In this diagram the center of the circle is $O$, the radius is $a$ inches, chord $EF$ is parallel to chord $CD, O, G, H, J$ are collinear, and $G$ is the midpoint of $CD$. Let $K$ (sq. in.) represent the area of trapezoid $CDFE$ and let $R$ (sq. in.) represent the area of rectangle $ELMF$. Then, as $CD$ and $EF$ are translated upward so that $OG$ increases toward the value $a$, while $JH$ always equals $HG$, the ratio $K:R$ become arbitrarily close to: [asy] size((270)); draw((0,0)--(10,0)..(5,5)..(0,0)); draw((5,0)--(5,5)); draw((9,3)--(1,3)--(1,1)--(9,1)--cycle); draw((9.9,1)--(.1,1)); label("O", (5,0), S); label("a", (7.5,0), S); label("G", (5,1), SE); label("J", (5,5), N); label("H", (5,3), NE); label("E", (1,3), NW); label("L", (1,1), S); label("C", (.1,1), W); label("F", (9,3), NE); label("M", (9,1), S); label("D", (9.9,1), E); [/asy] $\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ \sqrt{2} \qquad\textbf{(D)}\ \frac{1}{\sqrt{2}}+\frac{1}{2} \qquad\textbf{(E)}\ \frac{1}{\sqrt{2}}+1$

Two circles $O_1$ and $O_2$ intersect each other at $M$ and $N$. The common tangent to two circles nearer to $M$ touch $O_1$ and $O_2$ at $A$ and $B$ respectively. Let $C$ and $D$ be the reflection of $A$ and $B$ respectively with respect to $M$. The circumcircle of the triangle $DCM$ intersect circles $O_1$ and $O_2$ respectively at points $E$ and $F$ (both distinct from $M$). Show that the circumcircles of triangles $MEF$ and $NEF$ have same radius length.

Let $PRUS$ be a trapezium such that $\angle PSR = 2\angle QSU$ and $\angle SPU = 2 \angle UPR$. Let $Q$ and $T$ be on $PR$ and $SU$ respectively such that $SQ$ and $PU$ bisect $\angle PSR$ and $\angle SPU$ respectively. Let $PT$ meet $SQ$ at $E$. The line through $E$ parallel to $SR$ meets $PU$ in $F$ and the line through $E$ parallel to $PU$ meets $SR$ in $G$. Let $FG$ meet $PR$ and $SU$ in $K$ and $L$ respectively. Prove that $KF$ = $FG$ = $GL$.