A rectangle intersects a circle as shown: $AB=4$, $BC=5$, and $DE=3$. Then $EF$ equals: [asy]size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair D=origin, E=(3,0), F=(10,0), G=(12,0), H=(12,1), A=(0,1), B=(4,1), C=(9,1), O=circumcenter(B,C,F); draw(D--G--H--A--cycle); draw(Circle(O, abs(O-C))); label("$A$", A, NW); label("$B$", B, NW); label("$C$", C, NE); label("$D$", D, SW); label("$E$", E, SE); label("$F$", F, SW); label("4", (2,0.85), N); label("3", D--E, S); label("5", (6.5,0.85), N); [/asy] $\mathbf{(A)}\; 6\qquad \mathbf{(B)}\; 7\qquad \mathbf{(C)}\; \frac{20}3\qquad \mathbf{(D)}\; 8\qquad \mathbf{(E)}\; 9$

Let $ABC$ be a triangle such that $|AB|=13 , |BC|=12$ and $|CA|=5$. Let the angle bisectors of $A$ and $B$ intersect at $I$ and meet the opposing sides at $D$ and $E$, respectively. The line passing through $I$ and the midpoint of $[DE]$ meets $[AB]$ at $F$. What is $|AF|$? $ \textbf{(A)}\ \dfrac{3}{2} \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ \dfrac{5}{2} \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ \dfrac{7}{2} $

A trapezium has parallel sides of length equal to $a$ and $b$ ($a <b$), and the distance between the parallel sides is the altitude $h$. The extensions of the non-parallel lines intersect at a point that is a vertex of two triangles that have as sides the parallel sides of the trapezium. Express the areas of the triangles as functions of $a,b$ and $h$.

Let $ABCD$ be a trapezoid with $AD \parallel BC$, $AD < BC$. Let $E$ be a point on the side $AB$ and $F$ be point on the side $CD$. The circle $(AEF)$ intersects the segment $AD$ again at $A_1$ and the circle $(CEF)$ intersects these segment $BC$ again at $C_1$. Prove that the lines $A_1 C_1$, $BD$ and $EF$ are concurrent. [i]Proposed by A. Kuznetsov[/i]

Figure $ABCD$ is a trapezoid with $AB || DC, AB = 5, BC = 3 \sqrt 2, \measuredangle BCD = 45^\circ$, and $\measuredangle CDA = 60^\circ$. The length of $DC$ is $\textbf{(A) }7 + \frac{2}{3} \sqrt{3}\qquad \textbf{(B) }8\qquad \textbf{(C) }9 \frac{1}{2}\qquad \textbf{(D) }8 + \sqrt 3\qquad \textbf{(E) }8 + 3 \sqrt 3$

Suppose $ABCD$ is an isosceles trapezoid in which $\overline{AB}\parallel\overline{CD}$. Two mutually externally tangent circles $\omega_1$ and $\omega_2$ are inscribed in $ABCD$ such that $\omega_1$ is tangent to $\overline{AB}$,$\overline{BC}$, and $\overline{CD}$ while $\omega_2$ is tangent to $\overline{AB}$, $\overline{DA}$, and $\overline{CD}$. Given that $AB=1$, $CD=6$, compute the radius of either circle.

Given the circle $c(O,R)$ (with center $O$ and radius $R$), one diameter $AB$ and midpoint $C$ of the arc $AB$. Consider circle $c_1(K,KO)$, where center $K$ lies on the segment $OA$, and consider the tangents $CD,CO$ from the point $C$ to circle $c_1(K,KO)$. Line $KD$ intersects circle $c(O,R)$ at points $E$ and $Z$ (point $E$ lies on the semicircle that lies also point $C$). Lines $EC$ and $CZ$ intersects $AB$ at points $N$ and $M$ respectively. Prove that quadrilateral $EMZN$ is an isosceles trapezoid, inscribed in a circle whose center lie on circle $c(O,R)$.

In a triangle $ABC,$ the internal and external bisectors of the angle $A$ intersect the line $BC$ at $D$ and $E$ respectively. The line $AC$ meets the circle with diameter $DE$ again at $F.$ The tangent line to the circle $ABF$ at $A$ meets the circle with diameter $DE$ again at $G.$ Show that $AF = AG.$

The points $A, B, C, D$ lie in this order on a circle $o$. The point $S$ lies inside $o$ and has properties $\angle SAD=\angle SCB$ and $\angle SDA= \angle SBC$. Line which in which angle bisector of $\angle ASB$ in included cut the circle in points $P$ and $Q$. Prove $PS =QS$.

Let $K$ and $L$ be two points on the arcs $AB$ and $BC$ of the circumcircle of a triangle $ABC$, respectively, such that $KL\parallel AC$. Show that the incenters of triangles $ABK$ and $CBL$ are equidistant from the midpoint of the arc $ABC$ of the circumcircle of triangle $ABC$.

Let $ABCD$ be a right-angled trapezoid and $M$ be the midpoint of its greater lateral side $CD$. Circumcircles $\omega_{1}$ and $\omega_{2}$ of triangles $BCM$ and $AMD$ meet for the second time at point $E$. Let $ED$ meet $\omega_{1}$ at point $F$, and $FB$ meet $AD$ at point $G$. Prove that $GM$ bisects angle $BGD$.

A rhombus $ABCD$ with $\angle BAD=60^{\circ}$ is given. Points $E$ on side $AB$ and $F$ on side $AD$ are such that $\angle ECF=\angle ABD$. Lines $CE$ and $CF$ respectively meet line $BD$ at $P$ and $Q$. Prove that $\frac{PQ}{EF}=\frac{AB}{BD}$.

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

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.

Let $s=\frac{AB+BC+AC}{2}$ be half-perimeter of triangle $ABC$. Let $L$ and $N$be a point's on ray's $AB$ and $CB$, for which $AL=CN=s$. Let $K$ is point, symmetric of point $B$ by circumcenter of $ABC$. Prove, that perpendicular from $K$ to $NL$ passes through incenter of $ABC$. Solution for problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&t=365714&p=2011659#p2011659]here[/url]

Two circles $ C_1,C_2$ with different radii are given in the plane, they touch each other externally at $ T$. Consider any points $ A\in C_1$ and $ B\in C_2$, both different from $ T$, such that $ \angle ATB \equal{} 90^{\circ}$. (a) Show that all such lines $ AB$ are concurrent. (b) Find the locus of midpoints of all such segments $ AB$.

Circles $ C_1$ and $ C_2$ with centers at $ O_1$ and $ O_2$ respectively meet at points $ A$ and $ B$. The radii $ O_1B$ and $ O_2B$ meet $ C_1$ and $ C_2$ at $ F$ and$ E$. The line through $ B$ parallel to $ EF$ intersects $ C_1$ again at $ M$ and $ C_2$ again at $ N$. Prove that $ MN \equal{} AE \plus{} AF$.

The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with $ 9$ trapezoids, let $ x$ be the angle measure in degrees of the larger interior angle of the trapezoid. What is $ x$? [asy]unitsize(4mm); defaultpen(linewidth(.8pt)); int i; real r=5, R=6; path t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0); for(i=0; i<9; ++i) { draw(rotate(20*i)*t); } draw((-r,0)--(R+1,0)); draw((-R,0)--(-R-1,0));[/asy]$ \textbf{(A)}\ 100 \qquad \textbf{(B)}\ 102 \qquad \textbf{(C)}\ 104 \qquad \textbf{(D)}\ 106 \qquad \textbf{(E)}\ 108$

The coordinates of the vertices of isosceles trapezoid $ABCD$ are all integers, with $A=(20,100)$ and $D=(21,107).$ The trapezoid has no horizontal or vertical sides, and $\overline{AB}$ and $\overline{CD}$ are the only parallel sides. The sum o f the absolute values of all possible slopes for $\overline{AB}$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

In triangle $ABC$, $\angle B=2\angle C$ and $\angle A>90^\circ$. Let $D$ be the point on the line $AB$ such that $CD$ is perpendicular to $AC$, and let $M$ be the midpoint of $BC$. Prove that $\angle AMB=\angle DMC$.

What is the perimeter of trapezoid $ ABCD$? [asy]defaultpen(linewidth(0.8));size(3inch, 1.5inch); pair a=(0,0), b=(18,24), c=(68,24), d=(75,0), f=(68,0), e=(18,0); draw(a--b--c--d--cycle); draw(b--e); draw(shift(0,2)*e--shift(2,2)*e--shift(2,0)*e); label("30", (9,12), W); label("50", (43,24), N); label("25", (71.5, 12), E); label("24", (18, 12), E); label("$A$", a, SW); label("$B$", b, N); label("$C$", c, N); label("$D$", d, SE); label("$E$", e, S);[/asy] $ \textbf{(A)}\ 180\qquad\textbf{(B)}\ 188\qquad\textbf{(C)}\ 196\qquad\textbf{(D)}\ 200\qquad\textbf{(E)}\ 204 $

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

Let $ ABC$ be an acute triangle. Point $ D$ lies on side $ BC$. Let $ O_B, O_C$ be the circumcenters of triangles $ ABD$ and $ ACD$, respectively. Suppose that the points $ B, C, O_B, O_C$ lies on a circle centered at $ X$. Let $ H$ be the orthocenter of triangle $ ABC$. Prove that $ \angle{DAX} \equal{} \angle{DAH}$. [i]Zuming Feng.[/i]

In the accompanying figure, segments $AB$ and $CD$ are parallel, the measure of angle $D$ is twice the measure of angle $B$, and the measures of segments $AB$ and $CD$ are $a$ and $b$ respectively. Then the measure of $AB$ is equal to $\textbf{(A) }\dfrac{1}{2}a+2b\qquad\textbf{(B) }\dfrac{3}{2}b+\dfrac{3}{4}a\qquad\textbf{(C) }2a-b\qquad\textbf{(D) }4b-\dfrac{1}{2}a\qquad \textbf{(E) }a+b$ [asy] size(175); defaultpen(linewidth(0.8)); real r=50, a=4,b=2.5,c=6.25; pair A=origin,B=c*dir(r),D=(a,0),C=shift(b*dir(r))*D; draw(A--B--C--D--cycle); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,E); label("$D$",D,S); label("$a$",D/2,N); label("$b$",(C+D)/2,NW); //Credit to djmathman for the diagram[/asy]

Suppose that $ M$ is an arbitrary point on side $ BC$ of triangle $ ABC$. $ B_1,C_1$ are points on $ AB,AC$ such that $ MB = MB_1$ and $ MC = MC_1$. Suppose that $ H,I$ are orthocenter of triangle $ ABC$ and incenter of triangle $ MB_1C_1$. Prove that $ A,B_1,H,I,C_1$ lie on a circle.