In an acute angled triangle $ABC$, $\angle A = 30^{\circ}$, $H$ is the orthocenter, and $M$ is the midpoint of $BC$. On the line $HM$, take a point $T$ such that $HM = MT$. Show that $AT = 2 BC$.

On each side of a parallelogram an arbitrary point is chosen. Each pair of chosen points on neighbouring sides (i.e. sides with a common vertex) are connected by a line segment. Prove that the centres of the circumscribed circles of the four triangles so created are themselves vertices of a parallelogram. (ED Kulanin)

A triangle $ ABC$ has an obtuse angle at $ B$. The perpindicular at $ B$ to $ AB$ meets $ AC$ at $ D$, and $ |CD| \equal{} |AB|$. Prove that $ |AD|^2 \equal{} |AB|.|BC|$ if and only if $ \angle CBD \equal{} 30^\circ$.

Let $O$ be the center of the circumcircle $\omega$ of an acute-angle triangle $ABC$. A circle $\omega_1$ with center $K$ passes through $A$, $O$, $C$ and intersects $AB$ at $M$ and $BC$ at $N$. Point $L$ is symmetric to $K$ with respect to line $NM$. Prove that $BL \perp AC$.

The sides $AB$, $BC$, $CD$ and $DA$ of the quadrilateral $ABCD$ are respectively equal to the sides $A'B'$, $B'C'$, $C'D' $ and $D'A'$ of the quadrilateral $A'B'CD$' and it is known that $AB \parallel CD$ and $B'C' \parallel D'A'$. Prove that both quadrilaterals are parallelograms. (V Proizvolov, Moscow)

On the sides of a triangle $ABC$ are constructed similar triangles $ABD,BCE,CAF$ with $k=AD/DB=BE/EC=CF/FA$ and $\alpha=\angle ADB=\angle BEC=\angle CFA$. Prove that the midpoints of the segments $AC,BC,CD$ and $EF$ form a parallelogram with an angle $\alpha$ and two sides whose ratio is $k$.

Let $A_{1}A_{2}A_{3}$ be an acute triangle, and let $O$ and $H$ be its circumcenter and orthocenter, respectively. For $1\leq i \leq 3$, points $P_{i}$ and $Q_{i}$ lie on lines $OA_{i}$ and $A_{i+1}A_{i+2}$ (where $A_{i+3}=A_{i}$), respectively, such that $OP_{i}HQ_{i}$ is a parallelogram. Prove that \[\frac{OQ_{1}}{OP_{1}}+\frac{OQ_{2}}{OP_{2}}+\frac{OQ_{3}}{OP_{3}}\geq 3.\]

A rectangle formed by the lines of checkered paper is divided into figures of three kinds: isosceles right triangles (1) with base of two units, squares (2) with unit side, and parallelograms (3) formed by two sides and two diagonals of unit squares (figures may be oriented in any way). Prove that the number of figures of the third kind is even. [img]http://up.iranblog.com/Files7/dda310bab8b6455f90ce.jpg[/img]

A convex quadrilateral is given. Using a compass and a ruler construct a point such that its projections to the sidelines of this quadrilateral are the vertices of a parallelogram. (A. Zaslavsky)

In the parallelogram $ABCD$, the length of side $AB$ is half the length of side $BC$. The bisector of angle $\angle ABC$ intersects side $AD$ at point $K$ and diagonal $AC$ at point $L$. The bisector of angle $\angle ADC$ intersects the extension of side $AB$ at point $M$, with $B$ between $A$ and $M$. The line $ML$ intersects side $AD$ at point $F$. Calculate the ratio $\frac{AF}{AD}$.

A circumcircle of triangle $ABC$ meets the sides $AD$ and $CD$ of a parallelogram $ABCD$ at points $K$ and $L$ respectively. Let $M$ be the midpoint of arc $KL$ not containing $B$. Prove that $DM \perp AC$. by E.Bakaev

Consider an acute triangle $ABC$ of area $S$. Let $CD \perp AB$ ($D \in AB$), $DM \perp AC$ ($M \in AC$) and $DN \perp BC$ ($N \in BC$). Denote by $H_1$ and $H_2$ the orthocentres of the triangles $MNC$, respectively $MND$. Find the area of the quadrilateral $AH_1BH_2$ in terms of $S$.

In a convex quadrilateral $ABCD$, let $M$, $N$, $P$, and $Q$ be the midpoints of $AB$, $BC$, $CD$, and $DA$ respectively. If $MP$ and $NQ$ divide $ABCD$ in four quadrilaterals with the same area, prove that $ABCD$ is a parallelogram.

In a parallelogram $ABCD$ the trisectors of angles $A$ and $B$ are drawn. Let $O$ be the common points of the trisectors nearest to $AB$. Let $AO$ meet the second trisector of angle $B$ at point $A_1$, and let $BO$ meet the second trisector of angle $A$ at point $B_1$. Let $M$ be the midpoint of $A_1B_1$. Line $MO$ meets $AB$ at point $N$ Prove that triangle $A_1B_1N$ is equilateral.

Let $ABC$ be an acute-angled triangle with $AB<AC,$ and let $D$ be the foot of its altitude from$A,$ points $B'$ and $C'$ lie on the rays $AB$ and $AC,$ respectively , so that points $B',$ $C'$ and $D$ are collinear and points $B,$ $C,$ $B'$ and $C'$ lie on one circle with center $O.$ Prove that if $M$ is the midpoint of $BC$ and $H$ is the orthocenter of $ABC,$ then $DHMO$ is a parallelogram.

Let $ C_1$ and $ C_2$ be two congruent circles centered at $ O_1$ and $ O_2$, which intersect at $ A$ and $ B$. Take a point $ P$ on the arc $ AB$ of $ C_2$ which is contained in $ C_1$. $ AP$ meets $ C_1$ at $ C$, $ CB$ meets $ C_2$ at $ D$ and the bisector of $ \angle CAD$ intersects $ C_1$ and $ C_2$ at $ E$ and $ L$, respectively. Let $ F$ be the symmetric point of $ D$ with respect to the midpoint of $ PE$. Prove that there exists a point $ X$ satisfying $ \angle XFL \equal{} \angle XDC \equal{} 30^\circ$ and $ CX \equal{} O_1O_2$. [i] Author: Arnoldo Aguilar (El Salvador)[/i]

Let $ABCD$ be a trapezoid (quadrilateral with one pair of parallel sides) such that $AB < CD$. Suppose that $AC$ and $BD$ meet at $E$ and $AD$ and $BC$ meet at $F$. Construct the parallelograms $AEDK$ and $BECL$. Prove that $EF$ passes through the midpoint of the segment $KL$.

A trapezoid has side lengths $3, 5, 7,$ and $11$. The sum of all the possible areas of the trapezoid can be written in the form of $r_1 \sqrt{n_1} + r_2 \sqrt{n_2} + r_3$, where $r_1, r_2,$ and $r_3$ are rational numbers and $n_1$ and $n_2$ are positive integers not divisible by the square of a prime. What is the greatest integer less than or equal to \[r_1 + r_2 + r_3 + n_1 + n_2?\] $ \textbf{(A)}\ 57\qquad\textbf{(B)}\ 59\qquad\textbf{(C)}\ 61\qquad\textbf{(D)}\ 63\qquad\textbf{(E)}\ 65 $

Let $ABC$ be an acute triangle with $AB \neq AC$. Let $D$ be the foot of the altitude from $A$ and $\omega$ the circumcircle of the triangle. Let $\omega_1$ be the circle tangent to $AD$, $BD$ and $\omega$. Let $\omega_2$ be the circle tangent to $AD$, $CD$ and $\omega$. Let $\ell$ be the interior common tangent to both $\omega_1$ and $\omega_2$, different from $AD$. Prove that $\ell$ passes through the midpoint of $BC$ if and only if $2BC = AB + AC$.

$M$ is midpoint of $BC$.$P$ is an arbitary point on $BC$. $C_{1}$ is tangent to big circle.Suppose radius of $C_{1}$ is $r_{1}$ Radius of $C_{4}$ is equal to radius of $C_{1}$ and $C_{4}$ is tangent to $BC$ at P. $C_{2}$ and $C_{3}$ are tangent to big circle and line $BC$ and circle $C_{4}$. [img]http://aycu01.webshots.com/image/4120/2005120338156776027_rs.jpg[/img] Prove : \[r_{1}+r_{2}+r_{3}=R\] ($R$ radius of big circle)

1.14. Let P and Q be interior points of triangle ABC such that \ACP = \BCQ and \CAP = \BAQ. Denote by D;E and F the feet of the perpendiculars from P to the lines BC, CA and AB, respectively. Prove that if \DEF = 90, then Q is the orthocenter of triangle BDF.

Let $ABCD$ be a parallelogram with $\angle ABC=120^\circ$, $AB=16$ and $BC=10$. Extend $\overline{CD}$ through $D$ to $E$ so that $DE=4$. If $\overline{BE}$ intersects $\overline{AD}$ at $F$, then $FD$ is closest to $\textbf{(A) }1\qquad \textbf{(B) }2\qquad \textbf{(C) }3\qquad \textbf{(D) }4\qquad \textbf{(E) }5$ [asy] size(200); defaultpen(linewidth(0.8)); pair A=origin,B=(16,0),C=(26,10*sqrt(3)),D=(10,10*sqrt(3)),E=(0,10*sqrt(3)); draw(A--B--C--E--B--A--D); label("$A$",A,S); label("$B$",B,S); label("$C$",C,N); label("$D$",D,N); label("$E$",E,N); label("$F$",extension(A,D,B,E),W); label("$4$",(D+E)/2,N); label("$16$",(8,0),S); label("$10$",(B+C)/2,SE); [/asy]

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 $

A given convex pentagon $ ABCDE$ has the property that the area of each of five triangles $ ABC, BCD, CDE, DEA$, and $ EAB$ is unity [i](equal to 1)[/i]. Show that all pentagons with the above property have the same area, and calculate that area. Show, furthermore, that there are infinitely many non-congruent pentagons having the above area property.

The point $O$ is situated inside the parallelogram $ABCD$ such that $\angle AOB+\angle COD=180^{\circ}$. Prove that $\angle OBC=\angle ODC$.