On side \( AB \) of an isosceles trapezoid \( ABCD \) (\( AD \parallel BC \)), points \( E \) and \( F \) are chosen such that a circle can be inscribed in quadrilateral \( CDEF \). Prove that the circumcircles of triangles \( ADE \) and \( BCF \) are tangent to each other. [i]Proposed by Matthew Kurskyi[/i]

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.

$O$ is the intersection point of the diagonals of the trapezoid $ABCD$. A line passing through $C$ and a point symmetric to $B$ with respect to $O$, intersects the base $AD$ at the point $K$. Prove that $S_{AOK} = S_{AOB} + S_{DOK}$.

Let $ABCD$ to be a quadrilateral inscribed in a circle $\Gamma$. Let $r$ and $s$ to be the tangents to $\Gamma$ through $B$ and $C$, respectively, $M$ the intersection between the lines $r$ and $AD$ and $N$ the intersection between the lines $s$ and $AD$. After all, let $E$ to be the intersection between the lines $BN$ and $CM$, $F$ the intersection between the lines $AE$ and $BC$ and $L$ the midpoint of $BC$. Prove that the circuncircle of the triangle $DLF$ is tangent to $\Gamma$.

Let $H$ be the orthocenter of $\triangle ABC$. The points $A_{1} \not= A$, $B_{1} \not= B$ and $C_{1} \not= C$ lie, respectively, on the circumcircles of $\triangle BCH$, $\triangle CAH$ and $\triangle ABH$ and satisfy $A_{1}H=B_{1}H=C_{1}H$. Denote by $H_{1}$, $H_{2}$ and $H_{3}$ the orthocenters of $\triangle A_{1}BC$, $\triangle B_{1}CA$ and $\triangle C_{1}AB$, respectively. Prove that $\triangle A_{1}B_{1}C_{1}$ and $\triangle H_{1}H_{2}H_{3}$ have the same orthocenter.

Let $ABCD$ a convex quadrilateral with $AB=BC=CD$, with $AC$ not equal to $BD$ and $E$ be the intersection point of it's diagonals. Prove that $AE=DE$ if and only if $\angle BAD+\angle ADC = 120$.

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.

Let $ABCDEF$ be a hexagon with $AB||DE,\ BC||EF,\ CD||FA$ and in which the diagonals $AD,BE$ and $CF$ are congruent. Prove that the hexagon can be inscribed in a circle.

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] $\textbf{(A) }1\qquad \textbf{(B) }\dfrac{3}{2}\qquad \textbf{(C) }\dfrac{21}{13}\qquad \textbf{(D) }\dfrac{13}{8}\qquad \textbf{(E) }\dfrac{5}{3}\qquad$

In the figure, $AB \perp BC$, $BC \perp CD$, and $BC$ is tangent to the circle with center $O$ and diameter $AD$. In which one of the following cases is the area of $ABCD$ an integer? [asy] size(170); defaultpen(fontsize(10pt)+linewidth(.8pt)); pair O=origin, A=(-1/sqrt(2),1/sqrt(2)), B=(-1/sqrt(2),-1), C=(1/sqrt(2),-1), D=(1/sqrt(2),-1/sqrt(2)); draw(unitcircle); dot(O); draw(A--B--C--D--A); label("$A$",A,dir(A)); label("$B$",B,dir(B)); label("$C$",C,dir(C)); label("$D$",D,dir(D)); label("$O$",O,N); [/asy] $ \textbf{(A)}\ AB=3, CD=1\qquad\textbf{(B)}\ AB=5, CD=2\qquad\textbf{(C)}\ AB=7, CD=3\qquad\textbf{(D)}\ AB=9, CD=4\qquad\textbf{(E)}\ AB=11, CD=5 $

$A,B,C$ are on circle $\mathcal C$. $I$ is incenter of $ABC$ , $D$ is midpoint of arc $BAC$. $W$ is a circle that is tangent to $AB$ and $AC$ and tangent to $\mathcal C$ at $P$. ($W$ is in $\mathcal C$) Prove that $P$ and $I$ and $D$ are on a line.

Let $ABCD$ be a cyclic quadrilateral which is not a trapezoid and whose diagonals meet at $E$. The midpoints of $AB$ and $CD$ are $F$ and $G$ respectively, and $\ell$ is the line through $G$ parallel to $AB$. The feet of the perpendiculars from E onto the lines $\ell$ and $CD$ are $H$ and $K$, respectively. Prove that the lines $EF$ and $HK$ are perpendicular.

In the diagram, $ ABCD$ is a square, with $ U$ and $ V$ interior points of the sides $ AB$ and $ CD$ respectively. Determine all the possible ways of selecting $ U$ and $ V$ so as to maximize the area of the quadrilateral $ PUQV$. [img]http://i250.photobucket.com/albums/gg265/geometry101/CMO1992Number3.jpg[/img]

In a triangle $ABC$, let $D$ be the point on the segment $BC$ such that $AB+BD=AC+CD$. Suppose that the points $B$, $C$ and the centroids of triangles $ABD$ and $ACD$ lie on a circle. Prove that $AB=AC$.

Let $ABCD$ be a trapezoid with $AB \parallel CD$ and $AD = BD$. Let $M$ be the midpoint of $AB,$ and let $P \neq C$ be the second intersection of the circumcircle of $\triangle BCD$ and the diagonal $AC.$ Suppose that $BC = 27, CD = 25,$ and $AP = 10.$ If $MP = \tfrac {a}{b}$ for relatively prime positive integers $a$ and $b,$ compute $100a + b$.

Let $K$ be the measure of the area bounded by the $x$-axis, the line $x=8$, and the curve defined by \[f=\{(x,y)\,|\, y=x\text{ when }0\leq x\leq 5,\,y=2x-5\text{ when }5\leq x\leq 8\}.\] Then $K$ is: $\textbf{(A) }21.5\qquad \textbf{(B) }36.4\qquad \textbf{(C) }36.5\qquad \textbf{(D) }44\qquad$ $\textbf{ (E) }\text{less than 44 but arbitrarily close to it.}$

Prove that in a trapezoid with perpendicular diagonals, the product of the legs is at least as much as the product of the bases.

In $\triangle ABC$ with incenter $I$, $AB = 61$, $AC = 51$, and $BC=71$. The circumcircles of triangles $AIB$ and $AIC$ meet line $BC$ at points $D$ ($D \neq B$) and $E$ ($E \neq C$), respectively. Determine the length of segment $DE$. [i]James Tao[/i]

Let $H$ be the orthocentre of an acute triangle $ABC$. The line through $A$ perpendicular to $AC$ and the line through $B$ perpendicular to $BC$ intersect in $D$. The circle with centre $C$ through $H$ intersects the circumcircle of triangle $ABC$ in the points $E$ and $F$. Prove that $|DE| = |DF| = |AB|$.

A trapezoid $ABCD$ with $AB \parallel CD$ is inscribed in a circle $k$. Points $P$ and $Q$ are chose on the arc $ADCB$ in the order $A-P -Q-B$. Lines $CP$ and $AQ$ meet at $X$, and lines $BP$ and $DQ$ meet at $Y$. Show that points $P,Q,X, Y$ lie on a circle.

The quadrilateral $ABCD$ is an isosceles trapezoid, where $AB\parallel CD$. The trapezoid is inscribed in a circle with radius $R$ and center on side $AB$. Point $E$ lies on the circumscribed circle and is such that $\angle DAE = 90^o$. Given that $\frac{AE}{AB}=\frac34$, calculate the length of the sides of the isosceles trapezoid.

In a trapezoid $ABCD$ with $AB // CD, E$ is the midpoint of $BC$. Prove that if the quadrilaterals $ABED$ and $AECD$ are tangent, then the sides $a = AB, b = BC, c =CD, d = DA$ of the trapezoid satisfy the equalities $a+c = \frac{b}{3} +d$ and $\frac1a +\frac1c = \frac3b$ .

The polygon(s) formed by $y=3x+2$, $y=-3x+2$, and $y=-2$, is (are): $ \textbf{(A) }\text{An equilateral triangle}\qquad\textbf{(B) }\text{an isosceles triangle} \qquad\textbf{(C) }\text{a right triangle} \qquad$ $\textbf{(D) }\text{a triangle and a trapezoid}\qquad\textbf{(E) }\text{a quadrilateral} $

The trapezoid $ABCD$ satisfies $AB \parallel CD$, $AB = 70$, $AD = 32$ and $BC = 49$. We also know that $\angle ABC = 3 \angle ADC$. How long is the base $CD$?

In trapezium $ABCD$, $BC \perp AB$, $BC\perp CD$, and $AC\perp BD$. Given $AB=\sqrt{11}$ and $AD=\sqrt{1001}$. Find $BC$