Let's take a $ABCD$ rectangle of paper; the side $AB$ measures $5$ cm and the side $BC$ measures $9$ cm. We do three folds: 1.We take the $AB$ side on the $BC$ side and call $P$ the point on the $BC$ side that coincides with $A$. A right trapezoid $BCDQ$ is then formed. 2. We fold so that $B$ and $Q$ coincide. A $5$-sided polygon $RPCDQ$ is formed. 3. We fold again by matching $D$ with $C$ and $Q$ with $P$. A new right trapezoid $RPCS$. After making these folds, we make a cut perpendicular to $SC$ by its midpoint $T$, obtaining the right trapezoid $RUTS$. Calculate the area of the figure that appears as we unfold the last trapezoid $RUTS$.

Let $ABCD$ be a trapezium, $AD\parallel BC$, and let $E,F$ be points on the sides$AB$ and $CD$, respectively. The circumcircle of $AEF$ meets $AD$ again at $A_1$, and the circumcircle of $CEF$ meets $BC$ again at $C_1$. Prove that $A_1C_1,BD,EF$ are concurrent.

Point $M$ is the midpoint of base $AD$ of an isosceles trapezoid $ABCD$ with circumcircle $\omega$. The angle bisector of $ABD$ intersects $\omega$ at $K$. Line $CM$ meets $\omega$ again at $N$. From point $B$, tangents $BP, BQ$ are drawn to $(KMN)$. Prove that $BK, MN, PQ$ are concurrent. [i]A. Kuznetsov[/i]

Point $P$ is taken in the interior of the triangle $ABC$, so that \[\angle PAB = \angle PCB = \dfrac {1} {4} (\angle A + \angle C).\] Let $L$ be the foot of the angle bisector of $\angle B$. The line $PL$ meets the circumcircle of $\triangle APC$ at point $Q$. Prove that $QB$ is the angle bisector of $\angle AQC$. [i]Proposed by S. Berlov[/i]

Let $ ABCD$ be a trapezoid such that $ AB \parallel CD$ and assume that there are points $ E$ on the line outside the segment $ BC$ and $ F$ on the segment $ AD$ such that $ \angle DAE \equal{} \angle CBF$. Let $ I,J,K$ respectively be the intersection of line $ EF$ and line $ CD$, the intersection of line $ EF$ and line $ AB$, and the midpoint of segment $ EF$. Prove that $ K$ is on the circumcircle of triangle $ CDJ$ if and only if $ I$ is on the circumcircle of triangle $ ABK$. [i]Utari Wijayanti, Bandung[/i]

Let $ABC$ be a triangle and $X,Y,Z$ be the tangency points of its inscribed circle with the sides $BC, CA, AB$, respectively. Suppose that $C_1, C_2, C_3$ are circle with chords $YZ, ZX, XY$, respectively, such that $C_1$ and $C_2$ intersect on the line $CZ$ and that $C_1$ and $C_3$ intersect on the line $BY$. Suppose that $C_1$ intersects the chords $XY$ and $ZX$ at $J$ and $M$, respectively; that $C_2$ intersects the chords $YZ$ and $XY$ at $L$ and $I$, respectively; and that $C_3$ intersects the chords $YZ$ and $ZX$ at $K$ and $N$, respectively. Show that $I, J, K, L, M, N$ lie on the same circle.

The diagonals of a trapezoid $ ABCD$ intersect at point $ P$. Point $ Q$ lies between the parallel lines $ BC$ and $ AD$ such that $ \angle AQD \equal{} \angle CQB$, and line $ CD$ separates points $ P$ and $ Q$. Prove that $ \angle BQP \equal{} \angle DAQ$. [i]Author: Vyacheslav Yasinskiy, Ukraine[/i]

Given a triangle $ABC$, $\angle C= 60^o$. Construct a point $P$ on the side $AC$ and a point $Q$ on side $BC$ such that $ABQP$ is a trapezoid whose diagonals make an angle of $60^o$ with each other.

Let $ABCD$ be an isosceles trapezoid ($AD\parallel BC$), $\angle BAD = 80^o$, $\angle BDA = 60^o$. Point $P$ lies on $CD$ and $\angle PAD = 50^o$. Find $\angle PBC$

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 $AF = AG.$

Let $ABCD$ be a trapezium with $AB// DC, AB = b, AD = a ,a<b$ and $O$ the intersection point of the diagonals. Let $S$ be the area of the trapezium $ABCD$. Suppose the area of $\vartriangle DOC$ is $2S/9$. Find the value of $a/b$.

In the given circle, the diameter $\overline{EB}$ is parallel to $\overline{DC}$, and $\overline{AB}$ is parallel to $\overline{ED}$. The angles $AEB$ and $ABE$ are in the ratio $4:5$. What is the degree measure of angle $BCD$? [asy] unitsize(7mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real r=3; pair A=(-3cos(80),-3sin(80)); pair D=(3cos(80),3sin(80)), C=(-3cos(80),3sin(80)); pair O=(0,0), E=(-3,0), B=(3,0); path outer=Circle(O,r); draw(outer); draw(E--B); draw(E--A); draw(B--A); draw(E--D); draw(C--D); draw(B--C); pair[] ps={A,B,C,D,E,O}; dot(ps); label("$A$",A,N); label("$B$",B,NE); label("$C$",C,S); label("$D$",D,S); label("$E$",E,NW); label("$$",O,N);[/asy] $ \textbf{(A)}\ 120 \qquad \textbf{(B)}\ 125 \qquad \textbf{(C)}\ 130 \qquad \textbf{(D)}\ 135 \qquad \textbf{(E)}\ 140 $

Trapezoid $ABCD$ has parallel sides $\overline{AB}$ or length $33$ and $\overline{CD}$ of length $21$. The other two sides are of lengths $10$ and $14$. The angles at $A$ and $B$ are acute. What is the length of the shorter diagonal of $ABCD$? $ \textbf {(A) } 10\sqrt{6} \qquad \textbf {(B) } 25 \qquad \textbf {(C) } 8\sqrt{10} \qquad \textbf {(D) } 18\sqrt{2} \qquad \textbf {(E) } 26 $

Let $ABC$ be a triangle each of whose angles is greater than $30^{\circ}$. Suppose a circle centered with $P$ cuts segments $BC$ in $T,Q; CA$ in $K,L$ and $AB$ in $M,N$ such that they are on a circle in counterclockwise direction in that order.Suppose further $PQK,PLM,PNT$ are equilateral. Prove that: $a)$ The radius of the circle is $\frac{2abc}{a^2+b^2+c^2+4\sqrt{3}S}$ where $S$ is area. $b) a\cdot AP=b\cdot BP=c\cdot PC.$

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 $ABC$.