A convex quadrilateral $ABCD$ is inscribed in a circle with center $O$. The diagonals $AC$, $BD$ of $ABCD$ meet at $P$. Circumcircles of $\triangle ABP$ and $\triangle CDP$ meet at $P$ and $Q$ ($O,P,Q$ are pairwise distinct). Show that $\angle OQP=90^{\circ}$.

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 $PS =QS$.

Points $ A, B, C, D $ are vertices of an isosceles trapezoid, with $ \overline{AB} $ parallel to $ \overline{CD} $, $ AB = 1 $, $ CD = 2 $, and $ BC = 1 $. Point $ E $ is chosen uniformly and at random on $ \overline{CD} $, and let point $ F $ be the point on $ \overline{CD} $ such that $ EC = FD $. Let $ G $ denote the intersection of $ \overline{AE} $ and $ \overline{BF} $, not necessarily in the trapezoid. What is the probability that $ \angle AGB > 30^\circ $?

The trapezoid $ABCD$ has sides $AB$ and $CD$ that are parallel $\hat{DAB} = 6^{\circ}$ and $\hat{ABC} = 42^{\circ}$. Point $X$ lies on the side $AB$ , such that $\hat{AXD} = 78^{\circ}$ and $\hat{CXB} = 66^{\circ}$. The distance between $AB$ and $CD$ is $1$ unit . Prove that $AD + DX - (BC + CX) = 8$ units. (Heir of Ramanujan)

The diagram below shows two parallel rows with seven points in the upper row and nine points in the lower row. The points in each row are spaced one unit apart, and the two rows are two units apart. How many trapezoids which are not parallelograms have vertices in this set of $16$ points and have area of at least six square units? [asy] import graph; size(7cm); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; dot((-2,4),linewidth(6pt) + dotstyle); dot((-1,4),linewidth(6pt) + dotstyle); dot((0,4),linewidth(6pt) + dotstyle); dot((1,4),linewidth(6pt) + dotstyle); dot((2,4),linewidth(6pt) + dotstyle); dot((3,4),linewidth(6pt) + dotstyle); dot((4,4),linewidth(6pt) + dotstyle); dot((-3,2),linewidth(6pt) + dotstyle); dot((-2,2),linewidth(6pt) + dotstyle); dot((-1,2),linewidth(6pt) + dotstyle); dot((0,2),linewidth(6pt) + dotstyle); dot((1,2),linewidth(6pt) + dotstyle); dot((2,2),linewidth(6pt) + dotstyle); dot((3,2),linewidth(6pt) + dotstyle); dot((4,2),linewidth(6pt) + dotstyle); dot((5,2),linewidth(6pt) + dotstyle); [/asy]

In an isosceles trapezoid $ABCD$, the base $AB$ is twice as large as the base $CD$. Point $M$ is the midpoint of $AB$. It is known that the center of the circle inscribed in the triangle $MCB$ lies on the circle circumscribed around the triangle $MDC$. Find the angle $\angle MBC$. [img]https://cdn.artofproblemsolving.com/attachments/8/a/7af6a1d32c4e2affa49cb3eed9c10ba1e7ab71.png[/img]

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.

On the midline $MN$ of the trapezoid $ABCD$ ($AD\parallel BC$) the points $F$ and $G$ are chosen so that $\angle ABF =\angle CBG$. Prove that then $\angle BAF = \angle DAG$. (Dmitry Prokopenko)

Inside a circle of center $ O$ and radius $ r$, take two points $ A$ and $ B$ symmetrical about $ O$. We consider a variable point $ P$ on the circle and draw the chord $ \overline{PP'}\perp \overline{AP}$. Let $ C$ is the symmetric of $ B$ about $ \overline{PP'}$ ($ \overline{PP}'$ is the axis of symmetry) . Find the locus of point $ Q \equal{} \overline{PP'}\cap\overline{AC}$ when we change $ P$ in the circle.

Let $P$ and $Q$ be points on a circle $k$. A chord $AC$ of $k$ passes through the midpoint $M$ of $PQ$. Consider a trapezoid $ABCD$ inscribed in $k$ with $AB \parallel PQ \parallel CD$. Prove that the intersection point $X$ of $AD$ and $BC$ depends only on $k$ and $P,Q.$

An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $\mathcal{C}$ to the area of shaded region $\mathcal{B}$ is $11/5$. Find the ratio of shaded region $\mathcal{D}$ to the area of shaded region $\mathcal{A}$. [asy] defaultpen(linewidth(0.7)+fontsize(10)); for(int i=0; i<4; i=i+1) { fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray); } pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10); draw(B--A--C); fill((7.3,0)--(7.8,0)--(7.8,6)--(7.3,6)--cycle, white); clip(B--A--C--cycle); for(int i=0; i<9; i=i+1) { draw((i,1)--(i,6)); } label("$\mathcal{A}$", A+0.2*dir(-17), S); label("$\mathcal{B}$", A+2.3*dir(-17), S); label("$\mathcal{C}$", A+4.4*dir(-17), S); label("$\mathcal{D}$", A+6.5*dir(-17), S);[/asy]

Two parallel chords in a circle have lengths $10$ and $14$, and the distance between them is $6$. The chord parallel to these chords and midway between them is of length $\sqrt{a}$ where $a$ is [asy] // note: diagram deliberately not to scale -- azjps void htick(pair A, pair B, real r){ D(A--B); D(A-(r,0)--A+(r,0)); D(B-(r,0)--B+(r,0)); } size(120); pathpen = linewidth(0.7); pointpen = black+linewidth(3); real min = -0.6, step = 0.5; pair[] A, B; D(unitcircle); for(int i = 0; i < 3; ++i) { A.push(intersectionpoints((-9,min+i*step)--(9,min+i*step),unitcircle)[0]); B.push(intersectionpoints((-9,min+i*step)--(9,min+i*step),unitcircle)[1]); D(D(A[i])--D(B[i])); } MP("10",(A[0]+B[0])/2,N); MP("\sqrt{a}",(A[1]+B[1])/2,N); MP("14",(A[2]+B[2])/2,N); htick((B[1].x+0.1,B[0].y),(B[1].x+0.1,B[2].y),0.06); MP("6",(B[1].x+0.1,B[0].y/2+B[2].y/2),E);[/asy] $\textbf{(A)}\ 144 \qquad \textbf{(B)}\ 156 \qquad \textbf{(C)}\ 168 \qquad \textbf{(D)}\ 176 \qquad \textbf{(E)}\ 184$

Prove that if the two angles on the base of a trapezoid are different, then the diagonal starting from the smaller angle is longer than the other diagonal. [img]https://cdn.artofproblemsolving.com/attachments/7/1/77cf4958931df1c852c347158ff1e2bbcf45fd.png[/img]

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 $

In a quadrilateral $ABCD$ of area $1$, the parallel sides $BC$ and $AD$ are in the ratio $1 :2$ . $K$ is the midpoint of the diagonal $AC$ and $L$ is the point of intersection of the line $DK$ and the side $AB$. Determine the area of the quadrilateral $BCKL$ . (M G Sonkin)

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.

A trapezoid $ABCD$ with $BC // AD$ is given. The points $B'$ and $C'$ are symmetrical to $B$ and $C$ with respect to $CD$ and $AB$, respectively. Prove that the midpoint of the segment joining the circumcentres of $ABC'$ and $B'CD$ is equidistant from $A$ and $D$.

Given is the convex quadrilateral $ ABCD$. Assume that there exists a point $ P$ inside the quadrilateral for which the triangles $ ABP$ and $ CDP$ are both isosceles right triangles with the right angle at the common vertex $ P$. Prove that there exists a point $ Q$ for which the triangles $ BCQ$ and $ ADQ$ are also isosceles right triangles with the right angle at the common vertex $ Q$.

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} $

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?

The trapezium $[ABCD]$ has bases $[AB]$ and $[CD]$ (with $[AB]$ being the largest base). Knowing that $BC = 2 DA$ and that $\angle DAB + \angle ABC =120^o$ , determines the measure of $\angle DAB$.

Let $ABCD$ be an isosceles trapezoid with base $BC$ and $AD$. Suppose $\angle BDC=10^{\circ}$ and $\angle BDA=70^{\circ}$. Show that $AD^2=BC(AD+AB)$.

Let $E$ be a point outside the square $ABCD$ such that $m(\widehat{BEC})=90^{\circ}$, $F\in [CE]$, $[AF]\perp [CE]$, $|AB|=25$, and $|BE|=7$. What is $|AF|$? $ \textbf{(A)}\ 29 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 31 \qquad\textbf{(D)}\ 32 \qquad\textbf{(E)}\ 33 $

In triangle $ABC$ with $AB \neq AC$, points $N \in CA$, $M \in AB$, $P \in BC$, and $Q \in BC$ are chosen such that $MP \parallel AC$, $NQ \parallel AB$, $\frac{BP}{AB} = \frac{CQ}{AC}$, and $A, M, Q, P, N$ are concyclic. Find $\angle BAC$.

Given an isosceles trapezoid $ABCD$ with base $AB$. The diagonals $AC$ and $BD$ intersect at point $S$. Let $M$ the midpoint of $BC$ and the bisector of the angle $BSC$ intersect $BC$ at $N$. Prove that $\angle AMD = \angle AND$.