Let $ABCD$ be a parallelogram with at least an angle not equal to $90^o$ and $k$ the circumcircle of the triangle $ABC$. Let $E$ be the diametrically opposite point of $B$. Show that the circumcircle of the triangle $ADE$ and $k$ have the same radius.

Let $ABCD$ be a parallelogram and $H$ the Orthocentre of $\triangle{ABC}$. The line parallel to $AB$ through $H$ intersects $BC$ at $P$ and $AD$ at $Q$ while the line parallel to $BC$ through $H$ intersects $AB$ at $R$ and $CD$ at $S$. Show that $P$, $Q$, $R$ and $S$ are concyclic. [i](Swiss Mathematical Olympiad 2011, Final round, problem 8)[/i]

Consider the triangle $ABC$. Let $P, Q$ inside the angle $A$ such that $\angle BAP=\angle CAQ$ and $PBQC$ is a parallelogram. Show that $\angle ABP=\angle ACP.$

Let $ABCDEF$ be a convex hexagon. In the hexagon there is a point $K$, such that $ABCK,DEFK$ are both parallelograms. Prove that the three lines connecting $A,B,C$ to the midpoints of segments $CE,DF,EA$ meet at one point.

Let $ABC$ be acute-angled triangle . A circle $\omega_1(O_1,R_1)$ passes through points $B$ and $C$ and meets the sides $AB$ and $AC$ at points $D$ and $E$ ,respectively . Let $\omega_2(O_2,R_2)$ be the circumcircle of triangle $ADE$ . Prove that $O_1O_2$ is equal to the circumradius of triangle $ABC$ .

$ABCD$ is a parallelogram, line $DF$ is drawn bisecting $BC$ at $E$ and meeting $AB$ (extended) at $F$ from vertex $C$. Line $CH$ is drawn bisecting side $AD$ at $G$ and meeting $AB$ (extended) at $H$. Lines $DF$ and $CH$ intersect at $I$. If the area of parallelogram $ABCD$ is $x$, find the area of triangle $HFI$ in terms of $x$.

Let ABCD be a parallelogram, M the midpoint of AB and N the intersection of CD and the angle bisector of ABC. Prove that CM and BN are perpendicular iff AN is the angle bisector of DAB.

A infinite sequence $\{ a_n \}_{n \ge 0}$ of real numbers satisfy $a_n \ge n^2$. Suppose that for each $i, j \ge 0$ there exist $k, l$ with $(i,j) \neq (k,l)$, $l - k = j - i$, and $a_l - a_k = a_j - a_i$. Prove that $a_n \ge (n + 2016)^2$ for some $n$.

Let $ABCD$ be a convex quadrilateral whose sides $AD$ and $BC$ are not parallel. Suppose that the circles with diameters $AB$ and $CD$ meet at points $E$ and $F$ inside the quadrilateral. Let $\omega_E$ be the circle through the feet of the perpendiculars from $E$ to the lines $AB,BC$ and $CD$. Let $\omega_F$ be the circle through the feet of the perpendiculars from $F$ to the lines $CD,DA$ and $AB$. Prove that the midpoint of the segment $EF$ lies on the line through the two intersections of $\omega_E$ and $\omega_F$. [i]Proposed by Carlos Yuzo Shine, Brazil[/i]

Let $ABCD$ be a parallelogram. Point $E$ lies on segment $CD$ such that \[2\angle AEB=\angle ADB+\angle ACB,\] and point $F$ lies on segment $BC$ such that \[2\angle DFA=\angle DCA+\angle DBA.\] Let $K$ be the circumcenter of triangle $ABD$. Prove that $KE=KF$. [i]Merlijn Staps[/i]

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.

In parallelogram $ABCD$ of the accompanying diagram, line $DP$ is drawn bisecting $BC$ at $N$ and meeting $AB$ (extended) at $P$. From vertex $C$, line $CQ$ is drawn bisecting side $AD$ at $M$ and meeting $AB$ (extended) at $Q$. Lines $DP$ and $CQ$ meet at $O$. If the area of parallelogram $ABCD$ is $k$, then the area of the triangle $QPO$ is equal to [asy] size((400)); draw((0,0)--(5,0)--(6,3)--(1,3)--cycle); draw((6,3)--(-5,0)--(10,0)--(1,3)); label("A", (0,0), S); label("B", (5,0), S); label("C", (6,3), NE); label("D", (1,3), NW); label("P", (10,0), E); label("Q", (-5,0), W); label("M", (.5,1.5), NW); label("N", (5.65, 1.5), NE); label("O", (3.4,1.75)); [/asy] $ \textbf{(A)}\ k \qquad\textbf{(B)}\ \frac{6k}{5} \qquad\textbf{(C)}\ \frac{9k}{8} \qquad\textbf{(D)}\ \frac{5k}{4} \qquad\textbf{(E)}\ 2k $

Circles $ \omega_{1}$ and $ \omega_{2}$ intersect at points $ A$ and $ B.$ Points $ C$ and $ D$ are on circles $ \omega_{1}$ and $ \omega_{2},$ respectively, such that lines $ AC$ and $ AD$ are tangent to circles $ \omega_{2}$ and $ \omega_{1},$ respectively. Let $ I_{1}$ and $ I_{2}$ be the incenters of triangles $ ABC$ and $ ABD,$ respectively. Segments $ I_{1}I_{2}$ and $ AB$ intersect at $ E$. Prove that: $ \frac {1}{AE} \equal{} \frac {1}{AC} \plus{} \frac {1}{AD}$

Given is trapezoid $ABCD$, $M$ and $N$ being the midpoints of the bases of $AD$ and $BC$, respectively. a) Prove that the trapezoid is isosceles if it is known that the intersection point of perpendicular bisectors of the lateral sides belongs to the segment $MN$. b) Does the statement of point a) remain true if it is only known that the intersection point of perpendicular bisectors of the lateral sides belongs to the line $MN$?

Let $ABC$ be an acute triangle with incentre $I$ and $AB \neq AC$. Let lines $BI$ and $CI$ intersect the circumcircle of $ABC$ at $P \neq B$ and $Q \neq C$, respectively. Consider points $R$ and $S$ such that $AQRB$ and $ACSP$ are parallelograms (with $AQ \parallel RB, AB \parallel QR, AC \parallel SP$, and $AP \parallel CS$). Let $T$ be the point of intersection of lines $RB$ and $SC$. Prove that points $R, S, T$, and $I$ are concyclic.

$D$ is an arbitrary point lying on side $BC$ of $\triangle{ABC}$. Circle $\omega_1$ is tangent to segments $AD$ , $BD$ and the circumcircle of $\triangle{ABC}$ and circle $\omega_2$ is tangent to segments $AD$ , $CD$ and the circumcircle of $\triangle{ABC}$. Let $X$ and $Y$ be the intersection points of $\omega_1$ and $\omega_2$ with $BC$ respectively and take $M$ as the midpoint of $XY$. Let $T$ be the midpoint of arc $BC$ which does not contain $A$. If $I$ is the incenter of $\triangle{ABC}$, prove that $TM$ goes through the midpoint of $ID$.

A parallelogram $ABCD$ is given. On the diagonal BD, a point $P$ is selected such that $AP = BD$ is satisfied. Point $Q$ is the midpoint of segment $CP$. Prove that $\angle BQD = 90^o$. [img]https://cdn.artofproblemsolving.com/attachments/2/0/4bc69ec0330e2afa6b560c56da5dd783b16efb.png[/img] .

A parallelogram $ABCD$ is given ($AB \neq BC$). Points $E$ and $G$ are chosen on the line $\overline{CD}$ such that $\overline{AC}$ is the angle bisector of both angles $\angle EAD$ and $\angle BAG$. The line $\overline{BC}$ intersects $\overline{AE}$ and $\overline{AG}$ at $F$ and $H$, respectively. Prove that the line $\overline{FG}$ passes through the midpoint of $HE$. [i]Proposed by Mahdi Etesamifard[/i]

Let ABCD be a convex quadrilateral; P,Q, R,S are the midpoints of AB, BC, CD, DA respectively such that triangles AQR, CSP are equilateral. Prove that ABCD is a rhombus. Find its angles.

in an acute triangle $ABC$,$D$ is a point on $BC$,let $Q$ be the intersection of $AD$ and the median of $ABC$from $C$,$P$ is a point on $AD$,distinct from $Q$.the circumcircle of $CPD$ intersects $CQ$ at $C$ and $K$.prove that the circumcircle of $AKP$ passes through a fixed point differ from $A$.

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 $\triangle OAB$.

Given the parallelogram $ABCD$. The circle $S_1$ passes through the vertex $C$ and touches the sides $BA$ and $AD$ at points $P_1$ and $Q_1$, respectively. The circle $S_2$ passes through the vertex $B$ and touches the side $DC$ at points $P_2$ and $Q_2$, respectively. Let $d_1$ and $d_2$ be the distances from $C$ and $B$ to the lines $P_1Q_1$ and $P_2Q_2$, respectively. Find all possible values of the ratio $d_1:d_2$. [i](I. Voronovich)[/i]

Convex quadrilateral $ ABCD$ has $ AB\equal{}9$ and $ CD\equal{}12$. Diagonals $ AC$ and $ BD$ intersect at $ E$, $ AC\equal{}14$, and $ \triangle AED$ and $ \triangle BEC$ have equal areas. What is $ AE$? $ \textbf{(A)}\ \frac{9}{2}\qquad \textbf{(B)}\ \frac{50}{11}\qquad \textbf{(C)}\ \frac{21}{4}\qquad \textbf{(D)}\ \frac{17}{3}\qquad \textbf{(E)}\ 6$

Let $ABCD$ and $A^{\prime }B^{\prime}C^{\prime }D^{\prime }$ be two arbitrary parallelograms in the space, and let $M,$ $N,$ $P,$ $Q$ be points dividing the segments $AA^{\prime },$ $BB^{\prime },$ $CC^{\prime },$ $DD^{\prime }$ in equal ratios. [b]a.)[/b] Prove that the quadrilateral $MNPQ$ is a parallelogram. [b]b.)[/b] What is the locus of the center of the parallelogram $MNPQ,$ when the point $M$ moves on the segment $AA^{\prime }$ ? (Consecutive vertices of the parallelograms are labelled in alphabetical order.

Let $ A,B,C,$ and $ D$ denote four points in space and $ AB$ the distance between $ A$ and $ B$, and so on. Show that \[ AC^2\plus{}BD^2\plus{}AD^2\plus{}BC^2 \ge AB^2\plus{}CD^2.\]