Let $ABCD$ be a quadrilateral with parallel sides $AD$ and $BC$, $M$ and $N$ are the midpoints of its sides $AB$ and $CD$, respectively. The straight line $MN$ bisects the segment connecting the centers of the circumcircles of triangles $ABC$ and $ADC$. Prove that $ABCD$ is a parallelogram.

Let $ABCD$ be a square. Let $E, Z$ be points on the sides $AB, CD$ of the square respectively, such that $DE\parallel BZ$. Assume that the triangles $\triangle EAD, \triangle ZCB$ and the parallelogram $BEDZ$ have the same area. If the distance between the parallel lines $DE$ and $BZ$ is equal to $1$, determine the area of the square.

Consider $ \rho$ a semicircle of diameter $ AB$. A parallel to $ AB$ cuts the semicircle at $ C, D$ such that $ AD$ separates $ B, C$. The parallel at $ AD$ through $ C$ intersects the semicircle the second time at $ E$. Let $ F$ be the intersection point of the lines $ BE$ and $ CD$. The parallel through $ F$ at $ AD$ cuts $ AB$ in $ P$. Prove that $ PC$ is tangent to $ \rho$. [i]Author: Cosmin Pohoata[/i]

Let $P$ be a point in parallelogram $ABCD$ such that $$PA \cdot PC + PB \cdot PD = AB \cdot BC.$$ Prove that the reflections of $P$ over lines $AB$, $BC$, $CD$, and $DA$ are concyclic.

Let a triangle $ABC$ , $O$ it's circumcenter , $H$ ortocenter and $M$ the midpoint of $AH$. The perpendicular at $M$ to line $OM$ meets $AB$ and $AC$ at points $P$, respective $Q$. Prove that $MP=MQ$. Babis

On the side $AC$ of the triangle $ABC$ in the external side is constructed the parallelogram $ACDE$ . Let $O$ be the intersection point of its diagonals, $N$ and $K$ be midpoints of BC and BA respectively. Prove that lines $DK, EN$ and $BO$ intersect at one point.

Let $H$ be the orthocenter of the acute-angled triangle $ABC$. Inside the segment $BC$ arbitrary point $D$ is selected. Let $P$ be such that $ADPH$ is a parallelogram. Prove that $\angle BCP< \angle BHP$.

Laura won the local math olympiad and was awarded a "magical" ruler. With it, she can draw (as usual) lines in the plane, and she can also measure segments and replicate them anywhere in the plane; but she can also divide a segment into as many equal parts as she wishes; for instance, she can divide any segment into $17$ equal parts. Laura drew a parallelogram $ABCD$ and decided to try out her magical ruler; with it, she found the midpoint $M$ of side $CD$, and she extended $CB$ beyond $B$ to point $N$ so that segments $CB$ and $BN$ were equal in length. Unfortunately, her mischievous little brother came along and erased everything on Laura's picture except for points $A, M$, and $N$. Using Laura's magical ruler, help her reconstruct the original parallelogram $ABCD$: write down the steps that she needs to follow and prove why this will lead to reconstructing the original parallelogram $ABCD$.

Given convex quadrilateral $ABCD$. We construct the similar triangles $APB, BQC, CRD, DSA$ outside $ABCD$ so that $\angle PAB = \angle QBC = \angle RCD = \angle SDA, \angle PBA = \angle QCB = \angle RDC = \angle SAD$. Prove that if $PQRS$ is a parallelogram, so is $ABCD$.

In $ \triangle{ABC}$, we have $ AB\equal{}1$ and $ AC\equal{}2$. Side $ BC$ and the median from $ A$ to $ BC$ have the same length. What is $ BC$? $ \textbf{(A)}\ \frac{1\plus{}\sqrt2}{2} \qquad \textbf{(B)}\ \frac{1\plus{}\sqrt3}{2} \qquad \textbf{(C)}\ \sqrt2 \qquad \textbf{(D)}\ \frac{3}{2} \qquad \textbf{(E)}\ \sqrt3$

Let $ABC$ be a triangle with orthocenter $H$ and let $P$ be the second intersection of the circumcircle of triangle $AHC$ with the internal bisector of the angle $\angle BAC$. Let $X$ be the circumcenter of triangle $APB$ and $Y$ the orthocenter of triangle $APC$. Prove that the length of segment $XY$ is equal to the circumradius of triangle $ABC$.

Let $ABCD$ be a parallelogram with $AB=8$, $AD=11$, and $\angle BAD=60^\circ$. Let $X$ be on segment $CD$ with $CX/XD=1/3$ and $Y$ be on segment $AD$ with $AY/YD=1/2$. Let $Z$ be on segment $AB$ such that $AX$, $BY$, and $DZ$ are concurrent. Determine the area of triangle $XYZ$.

Let $ABCD$ be an inscribed quadrilateral in a circumference with center $O$ such that it lies inside $ABCD$ and $\angle{BAC} = \angle{ODA}$. Let $E$ be the intersection of $AC$ with $BD$. Lines $r$ and $s$ are drawn through $E$ such that $r$ is perpendicular to $BC$, and $s$ is perpendicular to $AD$. Let $P$ be the intersection of $r$ with $AD$, and $M$ the intersection of $s$ with $BC$. Let $N$ be the midpoint of $EO$. Prove that $M$, $N$, and $P$ lie on a line.

Point $ P$ is taken inside a right angle $ KLM$. A circle $ S_1$ with center $ O_1$ is tangent to the rays $ LK,LP$ of angle $ KLP$ at $ A,D$ respectively. A circle $ S_2$ with center $ O_2$ is tangent to the rays of angle $ MLP$, touching $ LP$ at $ B$. Suppose $ A,B,O_1$ are collinear. Let $ O_2D,KL$ meet at $ C$. Prove that $ BC$ bisects angle $ ABD$.

Consider parallelogram $ABCD$, with acute angle at $A$, $AC$ and $BD$ intersect at $E$. Circumscribed circle of triangle $ACD$ intersects $AB$, $BC$ and $BD$ at $K$, $L$ and $P$ (in that order). Then, circumscribed circle of triangle $CEL$ intersects $BD$ at $M$. Prove that: $$KD*KM=KL*PC$$

Let $l,m$ be two parallel lines in the plane. Let $P$ be a fixed point between them. Let $E,F$ be variable points on $l,m$ such that the angle $EPF$ is fixed to a number like $\alpha$ where $0<\alpha<\frac{\pi}2$. (By angle $EPF$ we mean the directed angle) Show that there is another point (not $P$) such that it sees the segment $EF$ with a fixed angle too.

Let $P$ be a point inside a triangle $ABC$ such that $\angle PAC= \angle PCB$. Let the projections of $P$ onto $BC$, $CA$, and $AB$ be $X,Y,Z$ respectively. Let $O$ be the circumcenter of $\triangle XYZ$, $H$ be the foot of the altitude from $B$ to $AC$, $N$ be the midpoint of $AC$, and $T$ be the point such that $TYPO$ is a parallelogram. Show that $\triangle THN$ is similar to $\triangle PBC$. [i]Proposed by Sammy Luo[/i]

In the plane, there are $n$ points ($n\ge 4$) where no 3 of them are collinear. Let $A(n)$ be the number of parallelograms whose vertices are those points with area $1$. Prove the following inequality: $A(n)\leq \frac{n^2-3n}{4}$ for all $n\ge 4$

Let $ABCD$ be a parallelogram. The circle with diameter $AC$ intersects the line $BD$ at points $P$ and $Q$. The perpendicular to the line $AC$ passing through the point $C$ intersects the lines $AB$ and $AD$ at points $X$ and $Y$, respectively. Prove that the points $P,Q,X$ and $Y$ lie on the same circle.

Let $ABCD$ be a parallelogram and a circle $k$ passes through $A, C$ and meets rays $AB, AD$ at $E, F$. If $BD, EF$ and the tangent at $C$ concur, show that $AC$ is diameter of $k$.

Let $ABCD$ be a tetrahedron with $AB=41$, $AC=7$, $AD=18$, $BC=36$, $BD=27$, and $CD=13$, as shown in the figure. Let $d$ be the distance between the midpoints of edges $AB$ and $CD$. Find $d^{2}$. [asy] pair C=origin, D=(4,11), A=(8,-5), B=(16,0); draw(A--B--C--D--B^^D--A--C); draw(midpoint(A--B)--midpoint(C--D), dashed); label("27", B--D, NE); label("41", A--B, SE); label("7", A--C, SW); label("$d$", midpoint(A--B)--midpoint(C--D), NE); label("18", (7,8), SW); label("13", (3,9), SW); pair point=(7,0); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D));[/asy]

Let $ABCD$ be a parallelogram. Suppose a line passing through $C$ and lying outside the parallelogram meets $AB$ and $AD$ produced at $E$ and $F$ respectively. Show that \[ AC^2 + CE \cdot CF = AB \cdot AE + AD \cdot AF . \]

A circle center $O$ is inscribed in the quadrilateral $ABCD$. $AB$ is parallel to and longer than $CD$ and has midpoint $M$. The line $OM$ meets $CD$ at $F$. $CD$ touches the circle at $E$. Show that $DE = CF$ iff $AB = 2CD$.

On the sides of the triangle $ABC$, three similar isosceles triangles $ABP \ (AP = PB)$, $AQC \ (AQ = QC)$, and $BRC \ (BR = RC)$ are constructed. The first two are constructed externally to the triangle $ABC$, but the third is placed in the same half-plane determined by the line $BC$ as the triangle $ABC$. Prove that $APRQ$ is a parallelogram.

Let $ABC$ be a triangle, the point $E$ is in the segment $AC$, the point $F$ is in the segment $AB$ and $P=BE\cap CF$. Let $D$ be a point such that $AEDF$ is a parallelogram, Prove that $D$ is in the side $BC$, if and only if, the triangle $BPC$ and the quadrilateral $AEPF$ have the same area.