Let $ABCD$ be a parallelogram, such that the point $M$ is the midpoint of the side $CD$ and lies on the bisector of the angle $\angle BAD$. Prove that $\angle AMB = 90^o$.

Inside the parallelogram $ABCD$ is a point $X$. Make a line that passes through point $X$ and divides the parallelogram into two parts whose areas differ from each other the most.

Consider the parallelogram $ ABCD, $ whose diagonals intersect at $ O. $ The bisector of the angle $ \angle DAC $ and that of $ \angle DBC $ intersect each other at $ T. $ Moreover, $ \overrightarrow{TD} +\overrightarrow{TC} =\overrightarrow{TO} . $ Find the angles of the triangle $ ABT. $

Let $p, q, r, s, t \in \mathbb{R}^{*}_{+}$ satisfying: i) $p^2 + pq + q^2 = s^2$ ii) $q^2 + qr + r^2 = t^2$ iii) $r^2 + rp + p^2 = s^2 - st + t^2$ Prove that \[\frac{s^2 - st + t^2}{s^2t^2} = \frac{r^2}{q^2t^2} + \frac{p^2}{q^2s^2} - \frac{pr}{q^2ts}\]

Consider an octagon with equal angles and with rational sides. Prove that it has a center of symmetry.

Let $ABCD$ be a convex quadrilateral with $AB$ parallel to $CD$. Let $P$ and $Q$ be the midpoints of $AC$ and $BD$, respectively. Prove that if $\angle ABP=\angle CBD$, then $\angle BCQ=\angle ACD$.

$AB$ is the diameter of semicircle $O$. $C$,$D$ are two arbitrary points on semicircle $O$. Point $P$ lies on line $CD$ such that line $PB$ is tangent to semicircle $O$ at $B$. Line $PO$ intersects line $CA$, $AD$ at point $E$, $F$ respectively. Prove that $OE$=$OF$.

Let $\triangle ABC$ be a triangle and $\omega$ its circumcircle. The tangent to $\omega$ through $B$ cuts the parallel to $BC$ through $A$ at $P$. The line $CP$ cuts the circumcircle of $\triangle ABP$ again in $Q$ and line $AQ$ cuts $\omega$ at $R$. Prove that $BQCR$ is parallelogram if and only if $AC=BC$.

Let $ABC$ be an acute angled scalene triangle with circumcentre $O$ and orthocentre $H.$ If $M$ is the midpoint of $BC,$ then show that $AO$ and $HM$ intersect on the circumcircle of $ABC.$

In a quadrilateral $ABCD$ with an incircle, $AB = CD; BC < AD$ and $BC$ is parallel to $AD$. Prove that the bisector of $\angle C$ bisects the area of $ABCD$.

On the diagonals $AC$ and $BD$ of the inscribed quadrilateral A$BCD$, the points $X$ and $Y$ are marked, respectively, so that the quadrilateral $ABXY$ is a parallelogram. Prove that the circumscribed circles of triangles $BXD$ and $CYA$ have equal radii. (Vyacheslav Yasinsky)

Let $ABCD$ be a trapezoid such that $|AC|=8$, $|BD|=6$, and $AD \parallel BC$. Let $P$ and $S$ be the midpoints of $[AD]$ and $[BC]$, respectively. If $|PS|=5$, find the area of the trapezoid $ABCD$.

On the sides of triangle $ABC$, isosceles right-angled triangles $AUB, CVB$, and $AWC$ are placed. These three triangles have their right angles at vertices $U, V$ , and $W$, respectively. Triangle $AUB$ lies completely inside triangle $ABC$ and triangles $CVB$ and $AWC$ lie completely outside $ABC$. See the figure. Prove that quadrilateral $UVCW$ is a parallelogram. [asy] import markers; unitsize(1.5 cm); pair A, B, C, U, V, W; A = (0,0); B = (2,0); C = (1.7,2.5); U = (B + rotate(90,A)*(B))/2; V = (B + rotate(90,C)*(B))/2; W = (C + rotate(90,A)*(C))/2; draw(A--B--C--cycle); draw(A--W, StickIntervalMarker(1,1,size=2mm)); draw(C--W, StickIntervalMarker(1,1,size=2mm)); draw(B--V, StickIntervalMarker(1,2,size=2mm)); draw(C--V, StickIntervalMarker(1,2,size=2mm)); draw(A--U, StickIntervalMarker(1,3,size=2mm)); draw(B--U, StickIntervalMarker(1,3,size=2mm)); draw(rightanglemark(A,U,B,5)); draw(rightanglemark(B,V,C,5)); draw(rightanglemark(A,W,C,5)); dot("$A$", A, S); dot("$B$", B, S); dot("$C$", C, N); dot("$U$", U, NE); dot("$V$", V, NE); dot("$W$", W, NW); [/asy]

(A.Zaslavsky) The circumradius of triangle $ ABC$ is equal to $ R$. Another circle with the same radius passes through the orthocenter $ H$ of this triangle and intersect its circumcirle in points $ X$, $ Y$. Point $ Z$ is the fourth vertex of parallelogram $ CXZY$. Find the circumradius of triangle $ ABZ$.

Let $ABC$ be a triangle. Prove that there exists a unique point $P$ for which one can find points $D$, $E$ and $F$ such that the quadrilaterals $APBF$, $BPCD$, $CPAE$, $EPFA$, $FPDB$, and $DPEC$ are all parallelograms. [i]Proposed by Lewis Chen[/i]

Let $M$ be the midpoint of (the smaller) arc $BC$ in circumcircle of triangle $ABC$. Suppose that the altitude drawn from $A$ intersects the circle at $N$. Draw two lines through circumcenter $O$ of $ABC$ paralell to $MB$ and $MC$, which intersect $AB$ and $AC$ at $K$ and $L$, respectively. Prove that $NK=NL$.

In a given plane $\varrho$ consider a convex quadrilateral $ABCD$ and denote $E=AC\cap BD.$ Moreover, consider a point $V\notin\varrho$. On rays $VA,VB,VC,VD$ find points $A',B',C',D'$ respectively such that $E,A',B',C',D'$ are coplanar and $A'B'C'D'$ is a parallelogram. Discuss conditions of solvability.

Let $ABCD$ be a parallelogram with an acute angle at $A$. Let $G$ be a point on the line $AB$, distinct from $B$, such that $|CG| = |CB|$. Let $H$ be a point on the line $BC$, distinct from $B$, such that $|AB| =|AH|$. Prove that triangle $DGH$ is isosceles. [asy] unitsize(1.5 cm); pair A, B, C, D, G, H; A = (0,0); B = (2,0); D = (0.5,1.5); C = B + D - A; G = reflect(A,B)*(C) + C - B; H = reflect(B,C)*(H) + A - B; draw(H--A--D--C--G); draw(interp(A,G,-0.1)--interp(A,G,1.1)); draw(interp(C,H,-0.1)--interp(C,H,1.1)); draw(D--G--H--cycle, dashed); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, E); dot("$D$", D, NW); dot("$G$", G, NE); dot("$H$", H, SE); [/asy]

Starting from a given cyclic quadrilateral $\mathcal{Q}_0$, a sequence of quadrilaterals is constructed so that $\mathcal{Q}_{k + 1}$ is the circumscribed quadrilateral of $\mathcal{Q}_k$ for $k = 0,1,\dots$. The sequence terminates when a quadrilateral is reached that is not cyclic. (The circumscribed quadrilateral of a cylic quadrilateral $ABCD$ has sides that are tangent to the circumcircle of $ABCD$ at $A$, $B$, $C$ and $D$.) Prove that the sequence always terminates, except when $\mathcal{Q}_0$ is a square.

In convex hexagon $AXBYCZ$, sides $AX$, $BY$ and $CZ$ are parallel to diagonals $BC$, $XC$ and $XY$, respectively. Prove that $\triangle ABC$ and $\triangle XYZ$ have the same area. [i]Proposed by Evan Chen[/i]

Let $ ABC$ be a triangle. The $ A$-excircle of triangle $ ABC$ has center $ O_{a}$ and touches the side $ BC$ at the point $ A_{a}$. The $ B$-excircle of triangle $ ABC$ touches its sidelines $ AB$ and $ BC$ at the points $ C_{b}$ and $ A_{b}$. The $ C$-excircle of triangle $ ABC$ touches its sidelines $ BC$ and $ CA$ at the points $ A_{c}$ and $ B_{c}$. The lines $ C_{b}A_{b}$ and $ A_{c}B_{c}$ intersect each other at some point $ X$. Prove that the quadrilateral $ AO_{a}A_{a}X$ is a parallelogram. [i]Remark.[/i] The $ A$[i]-excircle[/i] of a triangle $ ABC$ is defined as the circle which touches the segment $ BC$ and the extensions of the segments $ CA$ and $ AB$ beyound the points $ C$ and $ B$, respectively. The center of this circle is the point of intersection of the interior angle bisector of the angle $ CAB$ and the exterior angle bisectors of the angles $ ABC$ and $ BCA$. Similarly, the $ B$-excircle and the $ C$-excircle of triangle $ ABC$ are defined. [hide="Source of the problem"][i]Source of the problem:[/i] Theorem (88) in: John Sturgeon Mackay, [i]The Triangle and its Six Scribed Circles[/i], Proceedings of the Edinburgh Mathematical Society 1 (1883), pages 4-128 and drawings at the end of the volume.[/hide]

A convex quadrangle without parallel sides is given. For each triple of its vertices, a point is constructed that supplements this triple to a parallelogram, one of the diagonals of which coincides with the diagonal of the quadrangle. Prove that of the four points constructed, exactly one lies inside the original quadrangle.

Let $m$ and $n$ be positive integers. A sequence of points $(A_0,A_1,\ldots,A_n)$ on the Cartesian plane is called [i]interesting[/i] if $A_i$ are all lattice points, the slopes of $OA_0,OA_1,\cdots,OA_n$ are strictly increasing ($O$ is the origin) and the area of triangle $OA_iA_{i+1}$ is equal to $\frac{1}{2}$ for $i=0,1,\ldots,n-1$. Let $(B_0,B_1,\cdots,B_n)$ be a sequence of points. We may insert a point $B$ between $B_i$ and $B_{i+1}$ if $\overrightarrow{OB}=\overrightarrow{OB_i}+\overrightarrow{OB_{i+1}}$, and the resulting sequence $(B_0,B_1,\ldots,B_i,B,B_{i+1},\ldots,B_n)$ is called an [i]extension[/i] of the original sequence. Given two [i]interesting[/i] sequences $(C_0,C_1,\ldots,C_n)$ and $(D_0,D_1,\ldots,D_m)$, prove that if $C_0=D_0$ and $C_n=D_m$, then we may perform finitely many [i]extensions[/i] on each sequence until the resulting two sequences become identical.

The incircle of triangle $ABC$ touches sides $AB$;$BC$;$CA$ at $M$;$N$;$K$, respectively. The line through $A$ parallel to $NK$ meets $MN$ at $D$. The line through $A$ parallel to $MN$ meets $NK$ at $E$. Show that the line $DE$ bisects sides $AB$ and $AC$ of triangle $ABC$. [i]M. Sonkin[/i]

Point $M$ lies on the diagonal $BD$ of parallelogram $ABCD$ such that $MD = 3BM$. Lines $AM$ and $BC$ intersect in point $N$. What is the ratio of the area of triangle $MND$ to the area of parallelogram $ABCD$?