Found problems: 1049
An isosceles trapezoid $ABCD$ $(AB \parallel CD)$ is given. Points $E$ and $F$ lie on the
sides $BC$ and $AD$, and the points $M$ and $N$ lie on the segment $EF$ such that $DF = BE$ and
$FM = NE$. Let $K$ and $L$ be the foot of perpendicular lines from $M$ and $N$ to $AB$ and $CD$, respectively. Prove that $EKFL$ is a parallelogram.
[i]Proposed by Mahdi Etesamifard[/i]
The triangle $ ABC$ is given. On the extension of the side $ AB$ we construct the point $ R$ with $ BR \equal{} BC$, where $ AR > BR$ and on the extension of the side $ AC$ we construct the point $ S$ with $ CS \equal{} CB$, where $ AS > CS$. Let $ A_1$ be the point of intersection of the diagonals of the quadrilateral $ BRSC$.
Analogous we construct the point $ T$ on the extension of the side $ BC$, where $ CT \equal{} CA$ and $ BT > CT$ and on the extension of the side $ BA$ we construct the point $ U$ with $ AU \equal{} AC$, where $ BU > AU$. Let $ B_1$ be the point of intersection of the diagonals of the quadrilateral $ CTUA$.
Likewise we construct the point $ V$ on the extension of the side $ CA$, where $ AV \equal{} AB$ and $ CV > AV$ and on the extension of the side $ CB$ we construct the point $ W$ with $ BW \equal{} BA$ and $ CW > BW$. Let $ C_1$ be the point of intersection of the diagonals of the quadrilateral $ AVWB$.
Show that the area of the hexagon $ AC_1BA_1CB_1$ is equal to the sum of the areas of the triangles $ ABC$ and $ A_1B_1C_1$.
In a parallelogram $ABCD$ of surface area $60$ cm$^2$ , a line is drawn by $D$ that intersects $BC$ at $P$ and the extension of $AB$ at $Q$. If the area of the quadrilateral $ABPD$ is $46$ cm$^2$ , find the area of triangle $CPQ$.
Consider $\triangle XPQ$ and $\triangle YPQ$ such that $X$ and $Y$ are on the opposite sides of $PQ$ and the circumradius of $\triangle XPQ$ and the circumradius of $\triangle YPQ$ are the same. $I$ and $J$ are the incenters of $\triangle XPQ$ and $\triangle YPQ$ respectively. Let $M$ be the midpoint of $PQ$. Suppose $I, M, J$ are collinear. Prove that $XPYQ$ is a parallelogram.
Let $ABCDEF$ be a convex hexagon such that $BCEF$ is a parallelogram and $ABF$ an equilateral triangle. Given that $BC = 1, AD = 3, CD+DE = 2$, compute the area of $ABCDEF$
Let $ ABC$, $ l$ and $ P$ be arbitrary triangle, line and point. $ A',B',C'$ are reflections of $ A,B,C$ in point $ P$. $ A''$ is a point on $ B'C'$ such that $ AA''\parallel l$. $ B'',C''$ are defined similarly. Prove that $ A'',B'',C''$ are collinear.
The bisector of $\angle BAD$ of a parallelogram $ABCD$ meets $BC$ at $K$. The point $L$ lies on $AB$ such that $AL=CK$. The lines $AK$ and $CL$ meet at $M$. Let $(ALM)$ meet $AD$ after $D$ at $N$. Prove that $\angle CNL=90^{o}$
A triangle $ABC$ has acute angles at $A$ and $B$. Isosceles triangles $ACD$ and $BCE$ with bases $AC$ and $BC$ are constructed externally to triangle $ABC$ such that $\angle ADC = \angle ABC$ and $\angle BEC = \angle BAC$. Let $S$ be the circumcenter of $\triangle ABC$. Prove that the length of the polygonal line $DSE$ equals the perimeter of triangle $ABC$ if and only if $\angle ACB$ is right.
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]
In a ABCD cyclic quadrilateral 4 points K, L ,M, N are taken on AB , BC , CD and DA , respectively such that KLMN is a parallelogram. Lines AD, BC and KM have a common point. And also lines AB, DC and NL have a common point. Prove that KLMN is rhombus.
Consider $ \rho$ a semicircle of diameter $ AB$. A parallel to $ AB$ cuts the semicircle at $ C, D$ such that $ AD$ separates $ B, C$. The parallel at $ AD$ through $ C$ intersects the semicircle the second time at $ E$. Let $ F$ be the intersection point of the lines $ BE$ and $ CD$. The parallel through $ F$ at $ AD$ cuts $ AB$ in $ P$. Prove that $ PC$ is tangent to $ \rho$.
[i]Author: Cosmin Pohoata[/i]
It is given parallelogram $ABCD$. On it's sides $AB, BC, CD, DA$ are chosen points $E, F, G, H$ such that area of $EFGH$ is half of the area of $ABCD$. Show that at least one of the quadrilaterals $ABFH$ and $AEGD$ is parallelogram.
Two circles $ \omega_{1}$ and $ \omega_{2}$ intersect in points $ A$ and $ B$. Let $ PQ$ and $ RS$ be segments of common tangents to these circles (points $ P$ and $ R$ lie on $ \omega_{1}$, points $ Q$ and $ S$ lie on $ \omega_{2}$). It appears that $ RB\parallel PQ$. Ray $ RB$ intersects $ \omega_{2}$ in a point $ W\ne B$. Find $ RB/BW$.
[i]S. Berlov [/i]
Outside the parallelogram $ABCD$ on its sides $AB$ and $BC$ are constructed equilateral triangles $ABK$, and $BCM$. Prove that the triangle $KMD$ is equilateral.
Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$.
[i]Proposed by Vyacheslev Yasinskiy, Ukraine[/i]
Let $ABC$ and $XYZ$ be two triangles. Define
\[A_1=BC\cap ZX, A_2=BC\cap XY,\]\[B_1=CA\cap XY, B_2=CA\cap YZ,\]\[C_1=AB\cap YZ, C_2=AB\cap ZX.\]
Hereby, the abbreviation $g\cap h$ means the point of intersection of two lines $g$ and $h$.
Prove that $\frac{C_1C_2}{AB}=\frac{A_1A_2}{BC}=\frac{B_1B_2}{CA}$ holds if and only if $\frac{A_1C_2}{XZ}=\frac{C_1B_2}{ZY}=\frac{B_1A_2}{YX}$.
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 $\frac{AF}{AD}$.
$ABCD$ is a trapezium such that $AB\parallel DC$ and $\frac{AB}{DC}=\alpha >1$. Suppose $P$ and $Q$ are points on $AC$ and $BD$ respectively, such that
\[\frac{AP}{AC}=\frac{BQ}{BD}=\frac{\alpha -1}{\alpha+1}\]
Prove that $PQCD$ is a parallelogram.
In the parallelolgram A$BCD$ with the center $S$, let $O$ be the center of the circle of the inscribed triangle $ABD$ and let $T$ be the touch point with the diagonal $BD$. Prove that the lines $OS$ and $CT$ are parallel.
Let $ABCD$ be a cyclic quadrilateral such that the triangles $BCD$ and $CDA$ are not equilateral. Prove that if the Simson line of $A$ with respect to $\triangle BCD$ is perpendicular to the Euler line of $BCD$, then the Simson line of $B$ with respect to $\triangle ACD$ is perpendicular to the Euler line of $\triangle ACD$.
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.
Let $ABCD$ be a parallelogram with the angle at $A$ obtuse. Let $P$ be a point on segment $BD$. The circle with center $P$ passing through $A$ cuts line $AD$ at $A$ and $Y$ and cuts line $AB$ at $A$ and $X$. Line $AP$ intersects $BC$ at $Q$ and $CD$ at $R$. Prove $\angle XPY = \angle XQY + \angle XRY$.
Let $ABCD$ be a parallelogram with $\angle BAD<90^o$ and $AB> BC$ . The angle bisector of $BAD$ intersects line $CD$ at point $P$ and line $BC$ at point $Q$. Prove that the center of the circle circumscirbed around the triangle $CPQ$ is equidistant from points $B$ and $D$.
Point $ M$ is given in the interior of parallelogram $ ABCD$, and the point $ N$ inside triangle $ AMD$ is chosen so that $ < MNA \plus{} < MCB \equal{} MND \plus{} < MBC \equal{} 180^0$. Prove that $ MN$ is parallel to $ AB$.
Let $ O$ be the circumcenter of acute triangle $ ABC$. Point $ P$ is in the interior of triangle $ AOB$. Let $ D,E,F$ be the projections of $ P$ on the sides $ BC,CA,AB$, respectively. Prove that the parallelogram consisting of $ FE$ and $ FD$ as its adjacent sides lies inside triangle $ ABC$.