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]

Let $ABCD$ be a parallelogram, $AC$ intersects $BD$ at $I$. Consider point $G$ inside $\triangle ABC$ that satisfy $\angle IAG=\angle IBG\neq 45^{\circ}-\frac{\angle AIB}{4}$. Let $E,G$ be projections of $C$ on $AG$ and $D$ on $BG$. The $E-$ median line of $\triangle BEF$ and $F-$ median line of $\triangle AEF$ intersects at $H$. $a)$ Prove that $AF,BE$ and $IH$ concurrent. Call the concurrent point $L$. $b)$ Let $K$ be the intersection of $CE$ and $DF$. Let $J$ circumcenter of $(LAB)$ and $M,N$ are respectively be circumcenters of $(EIJ)$ and $(FIJ)$. Prove that $EM,FN$ and the line go through circumcenters of $(GAB),(KCD)$ are concurrent.

Consider a parallelogram $ABCD$ where $AB>AD$. Let $X$ and $Y$, distinct from $B$, be points on the ray $BD^\rightarrow$ such that $CX=CB$ and $AY=AB$. Prove that $DX=DY$. Note: The notation $BD^\rightarrow$ denotes the ray originating from point $B$ passing through point $D$.

Let $ABCD$ be a parallelogram. Let $X \notin AC $ lie inside $ABCD$ so that $\angle AXB = \angle CXD = 90^ {\circ}$ and $\Omega$ denote the circumcircle of $AXC$. Consider a diameter $EF$ of $\Omega$ and assume neither $E, \ X, \ B$ nor $F, \ X, \ D$ are collinear. Let $K \neq X$ be an intersection point of circumcircles of $BXE$ and $DXF$ and $L \neq X$ be such point on $\Omega$ so that $\angle KXL = 90^{\circ}$. Prove that $AB = KL$.

Let $ABC$ a triangle and $P$ a point such that the triangles $PAB, PBC, PCA$ have the same area and the same perimeter. Prove that if: a) $P$ is in the interior of the triangle $ABC$ then $ABC$ is equilateral. b) $P$ is in the exterior of the triangle $ABC$ then $ABC$ is right angled triangle.

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 $S$.

Let $A$ and $B$ be two distinct points on the circle $\Gamma$, not diametrically opposite. The point $P$, distinct from $A$ and $B$, varies on $\Gamma$. Find the locus of the orthocentre of triangle $ABP$.

In parallelogram $ ABCD$, point $ M$ is on $ \overline{AB}$ so that $ \frac{AM}{AB} \equal{} \frac{17}{1000}$ and point $ N$ is on $ \overline{AD}$ so that $ \frac{AN}{AD} \equal{} \frac{17}{2009}$. Let $ P$ be the point of intersection of $ \overline{AC}$ and $ \overline{MN}$. Find $ \frac{AC}{AP}$.

On sides $AB, AC, BC$ are points $M, X, Y$, respectively, such that $AX=MX$; $BY=MY$. $K$, $L$ are midpoints of $AY$ and $BX$. $O$ is circumcenter of $ABC$, $O_1$, $O_2$ are symmetric with $O$ with respect to $K$ and $L$. Prove that $X, Y, O_1, O_2$ are concyclic.

Let two chords $AC$ and $BD$ of a circle $k$ meet at the point $K$, and let $O$ be the center of $k$. Let $M$ and $N$ be the circumcenters of triangles $AKB$ and $CKD$. Show that the quadrilateral $OMKN$ is a parallelogram.

The points of intersection of $ xy \equal{} 12$ and $ x^2 \plus{} y^2 \equal{} 25$ are joined in succession. The resulting figure is: $ \textbf{(A)}\ \text{a straight line} \qquad\textbf{(B)}\ \text{an equilateral triangle} \qquad\textbf{(C)}\ \text{a parallelogram}$ $ \textbf{(D)}\ \text{a rectangle} \qquad\textbf{(E)}\ \text{a square}$

In a parallelogram $ABCD$, the perpendiculars from $A$ to $BC$ and $CD$ meet the line segments $BC$ and $CD$ at the points $E$ and $F$ respectively. Suppose $AC = 37$ cm and $EF = 35$ cm. Let $H$ be the orthocentre of $\vartriangle AEF$. Find the length of $AH$ in cm. Show the steps in your calculations.

In a parallelogram $ABCD$ , $BD$ is the largest diagonal. By matching $B$ with $D$ by a bend, a regular pentagon is formed. Calculate the measures of the angles formed by the diagonal $BD$ with each of the sides of the parallelogram.

Given an oriented line $\Delta$ and a fixed point $A$ on it, consider all trapezoids $ABCD$ one of whose bases $AB$ lies on $\Delta$, in the positive direction. Let $E,F$ be the midpoints of $AB$ and $CD$ respectively. Find the loci of vertices $B,C,D$ of trapezoids that satisfy the following: [i](i) [/i] $|AB| \leq a$ ($a$ fixed); [i](ii) [/i] $|EF| = l$ ($l$ fixed); [i](iii)[/i] the sum of squares of the nonparallel sides of the trapezoid is constant. [hide="Remark"] [b]Remark.[/b] The constants are chosen so that such trapezoids exist.[/hide]

There are $\frac{k(k+1)}{2}+1$ points on the planes, some are connected by disjoint segments ( also point can not lies on segment, that connects two other points). It is true, that plane is divided to some parallelograms and one infinite region. What maximum number of segments can be drawn ? [i] A.Kupavski, A. Polyanski[/i]

The sides of an equilateral triangle $ABC$ are divided into $n$ equal parts $(n \geq 2) .$ For each point on a side, we draw the lines parallel to other sides of the triangle $ABC,$ e.g. for $n=3$ we have the following diagram: [asy] unitsize(150); defaultpen(linewidth(0.7)); int n = 3; /* # of vertical lines, including AB */ pair A = (0,0), B = dir(-30), C = dir(30); draw(A--B--C--cycle,linewidth(2)); dot(A,UnFill(0)); dot(B,UnFill(0)); dot(C,UnFill(0)); label("$A$",A,W); label("$C$",C,NE); label("$B$",B,SE); for(int i = 1; i < n; ++i) { draw((i*A+(n-i)*B)/n--(i*A+(n-i)*C)/n); draw((i*B+(n-i)*A)/n--(i*B+(n-i)*C)/n); draw((i*C+(n-i)*A)/n--(i*C+(n-i)*B)/n); } [/asy] For each $n \geq 2,$ find the number of existing parallelograms.

$ABCD$ is a convex quadrilateral. $K$, $L$, $M$, $N$ are the midpoints of the sides $AB$, $BC$, $CD$, $DA$. $BD$ bisects $KM$ at $Q$. $QA = QB = QC = QD$ , and$\frac{LK}{LM} = \frac{CD}{CB}$. Prove that $ABCD$ is a square

We choose a point on each side of a parallelogram $ABCD$, let these four points be $P, Q, R$ and $S$. Then we divide the parallelogram into several regions using line segments as shown in the figure. The areas of the grey regions are given, except for one (see the figure). Find the area of the region marked with a question mark. [img]https://cdn.artofproblemsolving.com/attachments/4/7/dbd009042dabdb2eafc8fc74960e9011038dae.png[/img]

$ABCD$ is a parallelogram. $g$ is a line passing $A$. Prove that the distance from $C$ to $g$ is either the sum or the difference of the distance from $B$ to $g$, and the distance from $D$ to $g$.

In trapezoid $ ABCD$ with bases $ AB$ and $ CD$, we have $ AB\equal{}52$, $ BC\equal{}12$, $ CD\equal{}39$, and $ DA\equal{}5$. The area of $ ABCD$ is [asy] pair A,B,C,D; A=(0,0); B=(52,0); C=(38,20); D=(5,20); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--D--cycle); label("$A$",A,S); label("$B$",B,S); label("$C$",C,N); label("$D$",D,N); label("52",(A+B)/2,S); label("39",(C+D)/2,N); label("12",(B+C)/2,E); label("5",(D+A)/2,W);[/asy] $ \text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260$

a) The triangle $ABC$ was turned around the centre of the circumscribed circle by the angle less than $180$ degrees and thus was obtained the triangle $A_1B_1C_1$. The corresponding segments $[AB]$ and $[A_1B_1]$ intersect in the point $C_2, [BC]$ and $[B_1C_1]$ -- $A_2, [AC]$ and $[A_1C_1]$ -- $B_2$. Prove that the triangle $A_2B_2C_2$ is similar to the triangle $ABC$. b) The quadrangle $ABCD$ was turned around the centre of the circumscribed circle by the angle less than $180$ degrees and thus was obtained the quadrangle $A_1B_1C_1D_1$. Prove that the points of intersection of the corresponding lines ( $(AB$) and $(A_1B_1), (BC)$ and $(B_1C_1), (CD)$ and $(C_1D_1), (DA)$ and $(D_1A_1)$ ) are the vertices of the parallelogram.

The sides of an equilateral triangle $ABC$ are divided into $n$ equal parts $(n \geq 2) .$ For each point on a side, we draw the lines parallel to other sides of the triangle $ABC,$ e.g. for $n=3$ we have the following diagram: [asy] unitsize(150); defaultpen(linewidth(0.7)); int n = 3; /* # of vertical lines, including AB */ pair A = (0,0), B = dir(-30), C = dir(30); draw(A--B--C--cycle,linewidth(2)); dot(A,UnFill(0)); dot(B,UnFill(0)); dot(C,UnFill(0)); label("$A$",A,W); label("$C$",C,NE); label("$B$",B,SE); for(int i = 1; i < n; ++i) { draw((i*A+(n-i)*B)/n--(i*A+(n-i)*C)/n); draw((i*B+(n-i)*A)/n--(i*B+(n-i)*C)/n); draw((i*C+(n-i)*A)/n--(i*C+(n-i)*B)/n); } [/asy] For each $n \geq 2,$ find the number of existing parallelograms.

Inside the parallelogram $ABCD$, point $E$ is chosen, such that $AE = DE$ and $\angle ABE = 90^o$. Point $F$ is the midpoint of the side $BC$ . Find the measure of the angle $\angle DFE$.

$M$ 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)

On the sides of a parallelogram, squares are constructed outwards. Prove that the centers of these squares are vertices of a square.