Found problems: 844
Let $M$, $N$ be the midpoints of diagonals $AC$, $BD$ of a right-angled trapezoid $ABCD$ ($\measuredangle A=\measuredangle D = 90^\circ$).
The circumcircles of triangles $ABN$, $CDM$ meet the line $BC$ in the points $Q$, $R$.
Prove that the distances from $Q$, $R$ to the midpoint of $MN$ are equal.
Trapezium $ABCD$, where $BC||AD$, is inscribed in a circle, $E$ is midpoint of the arc $AD$ of this circle not containing point $C$ . Let $F$ be the foot of the perpendicular drawn from $E$ on the line tangent to the circle at the point $C$ . Prove that $BC=2CF$.
Let ABCD be trapezium that is inscribed in circle (O) with larger edge BC. P is a point lying outer segment BC. PA cut (O) at N(that means PA isn't tangent of (O)), the circle with diameter PD intersect (O) at E, DE meet BC at N. Prove that MN always pass through a fixed point.
An isosceles trapezoid with bases $a$ and $c$ and altitude $h$ is given.
a) On the axis of symmetry of this trapezoid, find all points $P$ such that both legs of the trapezoid subtend right angles at $P$;
b) Calculate the distance of $p$ from either base;
c) Determine under what conditions such points $P$ actually exist. Discuss various cases that might arise.
On the midline $MN$ of the trapezoid $ABCD$ ($AD\parallel BC$) the points $F$ and $G$ are chosen so that $\angle ABF =\angle CBG$. Prove that then $\angle BAF = \angle DAG$.
(Dmitry Prokopenko)
An isosceles trapezoid $ABCD$ with bases $BC$ and $AD$ is given. Circles with centers $O_1$ and $O_2$ are inscribed in triangles $ABC$ and $ABD$. Prove that line $O_1O_2$ is perpendicular on $BC$.
In trapezoid $ABCD$, $AB < CD$, $AB\perp BC$, $AB\parallel CD$, and the diagonals $AC$, $BD$ are perpendicular at point $P$. There is a point $Q$ on ray $CA$ past $A$ such that $QD\perp DC$. If
\[\frac{QP} {AP}+\frac{AP} {QP} = \left( \frac{51}{14}\right)^4 - 2,\]then $\frac{BP} {AP}-\frac{AP}{BP}$ can be expressed in the form $\frac{m}{n}$ for relatively prime positive integers $m,n$. Compute $m+n$.
[i]Ray Li.[/i]
Let $ M$ and $ N$ be points on the side $ BC$ of triangle $ ABC$, with the point $ M$ lying on the segment $ BN$, such that $ BM \equal{} CN$. Let $ P$ and $ Q$ be points on the segments $ AN$ and $ AM$, respectively, such that $ \measuredangle PMC \equal{}\measuredangle MAB$ and $ \measuredangle QNB \equal{}\measuredangle NAC$. Prove that $ \measuredangle QBC \equal{}\measuredangle PCB$.
Let there be a trapezoid $ABCD$ with bases $AD$ and $BC$. Points $M$ and $P$ are on sides $AB$ and $CD$ such that $CM$ and $BP$ intersect in $N$ and the pentagon $AMNPD$ is cyclic. Prove that the triangle $ADN$ is isosceles.
In trapezoid $ABCD$, diagonal $AC$ is the bisector of angle $A$. Point $K$ is the midpoint of diagonal $AC$. It is known that $DC = DK$. Find the ratio of the bases $AD: BC$.
Let $ ABCD$ be a trapezoid with parallel sides $ AB > CD$. Points $ K$ and $ L$ lie on the line segments $ AB$ and $ CD$, respectively, so that $AK/KB=DL/LC$. Suppose that there are points $ P$ and $ Q$ on the line segment $ KL$ satisfying \[\angle{APB} \equal{} \angle{BCD}\qquad\text{and}\qquad \angle{CQD} \equal{} \angle{ABC}.\] Prove that the points $ P$, $ Q$, $ B$ and $ C$ are concyclic.
[i]Proposed by Vyacheslev Yasinskiy, Ukraine[/i]
Let $ABCD$ be an isosceles trapezoid, $AD=BC$, $AB \parallel CD$. The diagonals of the trapezoid intersect at the point $O$, and the point $M$ is the midpoint of the side $AD$. The circle circumscribed around the triangle $BCM$ intersects the side $AD$ at the point $K$. Prove that $OK \parallel AB$.
A trapezoid $ABCD$ with bases $AB$ and $CD$ is such that the circumcircle of the triangle $BCD$ intersects the line $AD$ in a point $E$, distinct from $A$ and $D$. Prove that the circumcircle oF the triangle $ABE$ is tangent to the line $BC$.
Let $M$ be the midpoint of (the smaller) arc $BC$ in circumcircle of triangle $ABC$. Suppose that the altitude drawn from $A$ intersects the circle at $N$. Draw two lines through circumcenter $O$ of $ABC$ paralell to $MB$ and $MC$, which intersect $AB$ and $AC$ at $K$ and $L$, respectively. Prove that $NK=NL$.
Let $ABCD$ be a convex quadrilateral with $AB$ parallel to $CD$. Let $P$ and $Q$ be the midpoints of $AC$ and $BD$, respectively. Prove that if $\angle ABP=\angle CBD$, then $\angle BCQ=\angle ACD$.
Let circles $ \Gamma $ and $ \omega $ are circumcircle and incircle of the triangle $ABC$, the incircle touches sides $BC,CA,AB$ at the points $A_1,B_1,C_1$. Let $A_2$ and $B_2$ lies the lines $A_1I$ and $B_1I$ ($A_1$ and $A_2$ lies different sides from $I$, $B_1$ and $B_2$ lies different sides from $I$) such that $IA_2=IB_2=R$. Prove that :
(a) $AA_2=BB_2=IO$;
(b) The lines $AA_2$ and $BB_2$ intersect on the circle $ \Gamma ;$
Let $ABCD$ be a isosceles trapezoid such that $AB || CD$ and all of its sides are tangent to a circle. $[AD]$ touches this circle at $N$. $NC$ and $NB$ meet the circle again at $K$ and $L$, respectively. What is $\dfrac {|BN|}{|BL|} + \dfrac {|CN|}{|CK|}$?
$
\textbf{(A)}\ 4
\qquad\textbf{(B)}\ 6
\qquad\textbf{(C)}\ 8
\qquad\textbf{(D)}\ 9
\qquad\textbf{(E)}\ 10
$
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 $KL$.
(V.Yasinsky, Ukraine) Suppose $ X$ and $ Y$ are the common points of two circles $ \omega_1$ and $ \omega_2$. The third circle $ \omega$ is internally tangent to $ \omega_1$ and $ \omega_2$ in $ P$ and $ Q$ respectively. Segment $ XY$ intersects $ \omega$ in points $ M$ and $ N$. Rays $ PM$ and $ PN$ intersect $ \omega_1$ in points $ A$ and $ D$; rays $ QM$ and $ QN$ intersect $ \omega_2$ in points $ B$ and $ C$ respectively. Prove that $ AB \equal{} CD$.
An equilateral triangle whose side is $ 2$ is divided into a triangle and a trapezoid by a line drawn parallel to one of its sides. If the area of the trapezoid equals one-half of the area of the original triangle, the length of the median of the trapezoid is:
$ \textbf{(A)}\ \frac{\sqrt{6}}{2} \qquad
\textbf{(B)}\ \sqrt{2} \qquad
\textbf{(C)}\ 2\plus{}\sqrt{2} \qquad
\textbf{(D)}\ \frac{2\plus{}\sqrt{2}}{2} \qquad
\textbf{(E)}\ \frac{2\sqrt{3}\minus{}\sqrt{6}}{2}$
Trapezium $ABCD$ and parallelogram $MBDK$ are located so that the sides of the parallelogram are parallel to the diagonals of the trapezoid (see fig.). Prove that the area of the gray part is equal to the sum of the areas of the black part.
(Yu. Blinkov)
[img]https://cdn.artofproblemsolving.com/attachments/b/9/cfff83b1b85aea16b603995d4f3d327256b60b.png[/img]
The vertices of hexagon $ABCDEF$ lie on a circle. Sides $AB = CD = EF = 6$, and sides $BC = DE = F A = 10$. The area of the hexagon is $m\sqrt3$. Find $m$.
Let $ABCD$ be a trapezium with bases $AB$ and $CD$ such that $AB = 2 CD$. From $A$ the line $r$ is drawn perpendicular to $BC$ and from $B$ the line $t$ is drawn perpendicular to $AD$. Let $P$ be the intersection point of $r$ and $t$. From $C$ the line $s$ is drawn perpendicular to $BC$ and from $D$ the line $u$ perpendicular to $AD$. Let $Q$ be the intersection point of $s$ and $u$. If $R$ is the intersection point of the diagonals of the trapezium, prove that points $P, Q$ and $R$ are collinear.
Let $ABC$, $AC \not= BC$, be an acute triangle with orthocenter $H$ and incenter $I$. The lines $CH$ and $CI$ meet the circumcircle of $\bigtriangleup ABC$ at points $D$ and $L$, respectively. Prove that $\angle CIH = 90^{\circ}$ if and only if $\angle IDL = 90^{\circ}$
Given $\triangle{ABC}$ with $\angle{B}=60^{\circ}$ and $\angle{C}=30^{\circ}$, let $P,Q,R$ be points on the sides $BA,AC,CB$ respectively such that $BPQR$ is an isosceles trapezium with $PQ \parallel BR$ and $BP=QR$.\\
Find the maximum possible value of $\frac{2[ABC]}{[BPQR]}$ where $[S]$ denotes the area of any polygon $S$.