A point $K$ is taken inside parallelogram $ABCD$ so that the midpoint of $AD$ is equidistant from $K$ and $C$, and the midpoint of $CD$ is equidistant form $K$ and $A$. Let $N$ be the midpoint of $BK$. Prove that the angles $NAK$ and $NCK$ are equal.

The ratio of the sides of a parallelogram is $\lambda>1$. Given $\lambda$, determine the maximum of the acute angle subtended by the diagonals of the parallelogram.

Suppose $ABCD$ is a parallelogram. Consider circles $w_1$ and $w_2$ such that $w_1$ is tangent to segments $AB$ and $AD$ and $w_2$ is tangent to segments $BC$ and $CD$. Suppose that there exists a circle which is tangent to lines $AD$ and $DC$ and externally tangent to $w_1$ and $w_2$. Prove that there exists a circle which is tangent to lines $AB$ and $BC$ and also externally tangent to circles $w_1$ and $w_2$. [i]Proposed by Ali Khezeli[/i]

In the parallelogram $CMNP$ extend the bisectors of angles $MCN$ and $PCN$ and intersect with extensions of sides PN and $MN$ at points $A$ and $B$, respectively. Prove that the bisector of the original angle $C$ of the the parallelogram is perpendicular to $AB$. [img]https://cdn.artofproblemsolving.com/attachments/f/3/fde8ef133758e06b1faf8bdd815056173f9233.png[/img]

In the plane there is a quadrilateral $ ABCD $ and a point $ M $. Construct a parallelogram with center $ M $ and its vertices lying on the lines $ AB $, $ BC $, $ CD $, $ DA $.

It is known that $ABCD$ is a parallelogram. The point $E$ is taken so that $BCED$ is a cyclic quadrilateral. Let $\ell$ be a line that passes through $A$, intersects the segment $DC$ at point $F$ and intersects the extension of the line $BC$ at $G$. Given $EF = EG = EC$. Prove that $\ell$ is the bisector of the angle $\angle BAD$.

Consider $\triangle OAB$ with acute angle $AOB$. Thorugh a point $M \neq O$ perpendiculars are drawn to $OA$ and $OB$, the feet of which are $P$ and $Q$ respectively. The point of intersection of the altitudes of $\triangle OPQ$ is $H$. What is the locus of $H$ if $M$ is permitted to range over a) the side $AB$; b) the interior of $\triangle OAB$.

Let $ABCD$ be a parallelogram and $P$ a point in the plane. The line $BP$ intersects the circumcircle of $ABC$ again at $X$ and the line $DP$ intersects the circumcircle of $DAC$ again at $Y$. Let $M$ be the midpoint of the side $AC$. The point $N$ lies on the circumcircle of $PXY$ so that $MN$ is a tangent to this circle. Prove that the segments $MN$ and $AM$ have the same length. [i]Proposed by David Anghel[/i]

Trapezoid $ ABCD$ has $ AD\parallel{}BC$, $ BD \equal{} 1$, $ \angle DBA \equal{} 23^{\circ}$, and $ \angle BDC \equal{} 46^{\circ}$. The ratio $ BC: AD$ is $ 9: 5$. What is $ CD$? $ \textbf{(A)}\ \frac {7}{9}\qquad \textbf{(B)}\ \frac {4}{5}\qquad \textbf{(C)}\ \frac {13}{15} \qquad \textbf{(D)}\ \frac {8}{9}\qquad \textbf{(E)}\ \frac {14}{15}$

On the sides $AB, BC, CD$ and $DA$ of the parallelogram $ABCD$ marked the points $M, N, K$ and $F$. respectively. Is it possible to determine, using only compass, whether the area of ​​the quadrilateral $MNKF$ is equal to half the area of ​​the parallelogram $ABCD$?

Let $ABCD$ be a rhombus with $\angle DAB = 60^o$. Let $K, L$ be points on its sides $AD$ and $DC$ and $M$ a point on the diagonal $AC$ such that $KDLM$ is a parallelogram. Prove that triangle $BKL$ is equilateral.

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.

Let $a,b,c,d$ are integers such that $(a,b)=(c,d)=1$ and $ad-bc=k>0$. Prove that there are exactly $k$ pairs $(x_{1},x_{2})$ of rational numbers with $0\leq x_{1},x_{2}<1$ for which both $ax_{1}+bx_{2},cx_{1}+dx_{2}$ are integers.

A quadrilateral $ABCD$ is inscribed into a circle with center $O.$ Points $P$ and $Q$ are opposite to $C$ and $D$ respectively. Two tangents drawn to that circle at these points meet the line $AB$ in points $E$ and $F.$ ($A$ is between $E$ and $B$, $B$ is between $A$ and $F$). The line $EO$ meets $AC$ and $BC$ in points $X$ and $Y$ respectively, and the line $FO$ meets $AD$ and $BD$ in points $U$ and $V$ respectively. Prove that $XV=YU.$

Let $N$ be a point on the longest side $AC$ of a triangle $ABC$. The perpendicular bisectors of $AN$ and $NC$ intersect $AB$ and $BC$ respectively in $K$ and $M$. Prove that the circumcenter $O$ of $\triangle ABC$ lies on the circumcircle of triangle $KBM$.

Given is a parallelogram $ABCD$, with $AB <AC <BC$. Points $E$ and $F$ are selected on the circumcircle $\omega$ of $ABC$ so that the tangenst to $\omega$ at these points pass through point $D$ and the segments $AD$ and $CE$ intersect. It turned out that $\angle ABF = \angle DCE$. Find the angle $\angle{ABC}$. A. Yakubov, S. Berlov

Let $ABCDE$ be a convex pentagon such that $BC \parallel AE,$ $AB = BC + AE,$ and $\angle ABC = \angle CDE.$ Let $M$ be the midpoint of $CE,$ and let $O$ be the circumcenter of triangle $BCD.$ Given that $\angle DMO = 90^{\circ},$ prove that $2 \angle BDA = \angle CDE.$ [i]Proposed by Nazar Serdyuk, Ukraine[/i]

Let a non-isosceles acute triangle ABC with the circumscribed cycle (O) and the orthocenter H. D, E, F are the reflection of O in the lines BC, CA and AB. a) $H_a$ is the reflection of H in BC, A' is the reflection of A at O and $O_a$ is the center of (BOC). Prove that $H_aD$ and OA' intersect on (O). b) Let X is a point satisfy AXDA' is a parallelogram. Prove that (AHX), (ABF), (ACE) have a comom point different than A

Given a parallelogram $ABCD$ with $AB<BC$, show that the circumcircles of the triangles $APQ$ share a second common point (apart from $A$) as $P,Q$ move on the sides $BC,CD$ respectively s.t. $CP=CQ$.

Let $ABCD$ be a quadrilateral (with non-perpendicular diagonals). The perpendicular from $A$ to $BC$ meets $CD$ at $K$. The perpendicular from $A$ to $CD$ meets $BC$ at $L$. The perpendicular from $C$ to $AB$ meets $AD$ at $M$. The perpendicular from $C$ to $AD$ meets $AB$ at $N$. 1. Prove that $KL$ is parallel to $MN$. 2. Prove that $KLMN$ is a parallelogram if $ABCD$ is cyclic.

Let the medians of the triangle $ABC$ meet at $G$. Let $D$ and $E$ be different points on the line $BC$ such that $DC=CE=AB$, and let $P$ and $Q$ be points on the segments $BD$ and $BE$, respectively, such that $2BP=PD$ and $2BQ=QE$. Determine $\angle PGQ$.

Let $ABCDE$ be a convex pentagon such that $BC \parallel AE,$ $AB = BC + AE,$ and $\angle ABC = \angle CDE.$ Let $M$ be the midpoint of $CE,$ and let $O$ be the circumcenter of triangle $BCD.$ Given that $\angle DMO = 90^{\circ},$ prove that $2 \angle BDA = \angle CDE.$ [i]Proposed by Nazar Serdyuk, Ukraine[/i]

The points $P$ and $Q$ lie on the sides $BC$ and $CD$ of the parallelogram $ABCD$ so that $BP = QD$. Show that the intersection point between the lines $BQ$ and $DP$ lies on the line bisecting $\angle BAD$.

Let $ ABC$ be a triangle, $ D$ be the midpoint of $ BC$, $ E$ be a point on segment $ AC$ such that $ BE\equal{}2AD$ and $ F$ is the intersection point of $ AD$ with $ BE$. If $ \angle DAC\equal{}60^{\circ}$, find the measure of the angle $ FEA$.

Equilateral triangles $ABE$ and $BCF$ are erected externally onthe sidess $AB$ and $BC$ of a parallelogram $ABCD$. Prove that $\vartriangle DEF$ is equilateral.