Found problems: 844
A trapezium $ ABCD$, in which $ AB$ is parallel to $ CD$, is inscribed in a circle with centre $ O$. Suppose the diagonals $ AC$ and $ BD$ of the trapezium intersect at $ M$, and $ OM \equal{} 2$.
[b](a)[/b] If $ \angle AMB$ is $ 60^\circ ,$ find, with proof, the difference between the lengths of the parallel sides.
[b](b)[/b] If $ \angle AMD$ is $ 60^\circ ,$ find, with proof, the difference between the lengths of the parallel sides.
[b][Weightage 17/100][/b]
Point $E$ is the midpoint of the base $AD$ of the trapezoid $ABCD$. Segments $BD$ and $CE$ intersect at point $F$. It is known that $AF$ is perpendicular to $BD$. Prove that $BC = FC$.
In the trapezoid $ABCD$ with bases $BC = a$, $AD = b$, the equality holds: $\angle BAC + \angle ACD = 180^o$. The straight line $AC$ intersects the common tangents to the circumcircles of the triangles $ABC$ and $ACD$ at the points Find $PQ$.
Given an isosceles trapezoid $ABCD$. The bisector of angle $B$ intersects the base $AD$ at point $L$. Prove that the center of the circle circumscribed around triangle $BLD$ lies on the circle circumscribed around the trapezoid.
(Yu. Blinkov)
Let $ABC$ be a triangle. Prove that there is a line $\ell$ (in the plane of triangle $ABC$) such that the intersection of the interior of triangle $ABC$ and the interior of its reflection $A'B'C'$ in $\ell$ has area more than $\frac23$ the area of triangle $ABC$.
Trapezoid $ABCD$ is an isosceles trapezoid with $AD=BC$. Point $P$ is the intersection of the diagonals $AC$ and $BD$. If the area of $\triangle ABP$ is $50$ and the area of $\triangle CDP$ is $72$, what is the area of the entire trapezoid?
[i]Proposed by David Altizio
Let $ ABCD$ be an isosceles trapezoid with $ AD \parallel BC$. Its diagonals $ AC$ and $ BD$ intersect at point $ M$. Points $ X$ and $ Y$ on the segment $ AB$ are such that $ AX \equal{} AM$, $ BY \equal{} BM$. Let $ Z$ be the midpoint of $ XY$ and $ N$ is the point of intersection of the segments $ XD$ and $ YC$. Prove that the line $ ZN$ is parallel to the bases of the trapezoid.
[i]Author: A. Akopyan, A. Myakishev[/i]
Let $ABC$ be a triangle such that $\angle ACB=2\angle ABC$. Let $D$ be the point on the side $BC$ such that $CD=2BD$. The segment $AD$ is extended to $E$ so that $AD=DE$. Prove that \[ \angle ECB+180^{\circ }=2\angle EBC. \]
Let $ABCD$ be an inscribed trapezoid such that the sides $[AB]$ and $[CD]$ are parallel. If $m(\widehat{AOD})=60^\circ$ and the altitude of the trapezoid is $10$, what is the area of the trapezoid?
The incircle of a triangle $ABC$ touches its sides $BC$, $CA$, $AB$ at the points $X$, $Y$, $Z$, respectively. Let $X^{\prime}$, $Y^{\prime}$, $Z^{\prime}$ be the reflections of these points $X$, $Y$, $Z$ in the external angle bisectors of the angles $CAB$, $ABC$, $BCA$, respectively. Show that $Y^{\prime}Z^{\prime}\parallel BC$, $Z^{\prime}X^{\prime}\parallel CA$ and $X^{\prime}Y^{\prime}\parallel AB$.
The line joining the midpoints of the diagonals of a trapezoid has length $3$. If the longer base is $97$, then the shorter base is:
$ \textbf{(A)}\ 94 \qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 89 $
The trapezium $[ABCD]$ has bases $[AB]$ and $[CD]$ (with $[AB]$ being the largest base). Knowing that $BC = 2 DA$ and that $\angle DAB + \angle ABC =120^o$ , determines the measure of $\angle DAB$.
In triangle $ABC$ with $AB>BC$, $BM$ is a median and $BL$ is an angle bisector. The line through $M$ and parallel to $AB$ intersects $BL$ at point $D$, and the line through $L$ and parallel to $BC$ intersects $BM$ at point $E$. Prove that $ED$ is perpendicular to $BL$.
A trapezoid with bases $a,b$ and altitude $h$ is circumscribed around a circl.. Prove that $h^2\le ab$.
In a trapezoid $ABCD$ with $AB // CD$, the diagonals $AC$ and $BD$ intersect at point $E$. Let $F$ and $G$ be the orthocenters of the triangles $EBC$ and $EAD$. Prove that the midpoint of $GF$ lies on the perpendicular from $E$ to $AB$.
On the trapezoid $ABCD$ , side $DA$ is perpendicular to the bases $AB$ and $CD$ . The base $AB$ measures $45$, the base $CD$ measures $20$ and the $BC$ side measures $65$. Let $P$ on the $BC$ side such that $BP$ measures $45$ and $M$ is the midpoint of $DA$. Calculate the measure of the $PM$ segment.
Consider the triangle $ ABC$ with $ m(\angle BAC \equal{} 90^\circ)$ and $ AC \equal{} 2AB$. Let $ P$ and $ Q$ be the midpoints of $ AB$ and $ AC$,respectively. Let $ M$ and $ N$ be two points found on the side $ BC$ such that $ CM \equal{} BN \equal{} x$. It is also known that $ 2S[MNPQ] \equal{} S[ABC]$. Determine $ x$ in function of $ AB$.
Let $ABC$ and $PQR$ be two triangles such that
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[b](a)[/b] $P$ is the mid-point of $BC$ and $A$ is the midpoint of $QR$.
[b](b)[/b] $QR$ bisects $\angle BAC$ and $BC$ bisects $\angle QPR$
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Prove that $AB+AC=PQ+PR$.
In the trapezoid $ABCD$ we know that $CD \perp BC, $ and $CD \perp AD .$ Circle $w$ with diameter $AB$ intersects $AD$ in points $A$ and $P,$ tangent from $P$ to $w$ intersects $CD$ at $M.$ The second tangent from $M$ to $w$ touches $w$ at $Q.$ Prove that midpoint of $CD$ lies on $BQ.$
The bisectors of trapezoid's angles form a quadrilateral with perpendicular diagonals. Prove that this trapezoid is isosceles.
Let $ABCD$ be a trapezoid with parallel sides $AB, CD$. $M,N$ lie on lines $AD, BC$ respectively such that $MN || AB$. Prove that $DC \cdot MA + AB \cdot MD = MN \cdot AD$.
On the midline of the isosceles trapezoid $ABCD$ ($BC \parallel AD$) find the point $K$, for which the sum of the angles $\angle DAK + \angle BCK$ will be the smallest.
The diagonals of a trapezoid $ ABCD$ intersect at point $ P$. Point $ Q$ lies between the parallel lines $ BC$ and $ AD$ such that $ \angle AQD \equal{} \angle CQB$, and line $ CD$ separates points $ P$ and $ Q$. Prove that $ \angle BQP \equal{} \angle DAQ$.
[i]Author: Vyacheslav Yasinskiy, Ukraine[/i]
In trapezoid $ABCD$, $AD \parallel BC$ and $\angle ABC + \angle CDA = 270^{\circ}$. Compute $AB^2$ given that $AB \cdot \tan(\angle BCD) = 20$ and $CD = 13$.
[i]Proposed by Lewis Chen[/i]
Let $ABCD$ be the isosceles trapezium with bases $AB$ and $CD$, such that $AC = BC$. The point $M$ is the midpoint of side $AD$. Prove that $\sphericalangle ACM = \sphericalangle CBD$.