Found problems: 3585
Let $ABCDE$ be a convex pentagon such that $BC \parallel AE,$ $AB = BC + AE,$ and $\angle ABC = \angle CDE.$ Let $M$ be the midpoint of $CE,$ and let $O$ be the circumcenter of triangle $BCD.$ Given that $\angle DMO = 90^{\circ},$ prove that $2 \angle BDA = \angle CDE.$
[i]Proposed by Nazar Serdyuk, Ukraine[/i]
Quadrilateral $APBQ$ is inscribed in circle $\omega$ with $\angle P = \angle Q = 90^{\circ}$ and $AP = AQ < BP$. Let $X$ be a variable point on segment $\overline{PQ}$. Line $AX$ meets $\omega$ again at $S$ (other than $A$). Point $T$ lies on arc $AQB$ of $\omega$ such that $\overline{XT}$ is perpendicular to $\overline{AX}$. Let $M$ denote the midpoint of chord $\overline{ST}$. As $X$ varies on segment $\overline{PQ}$, show that $M$ moves along a circle.
Given an acute triangle whose circumcenter is $O$.let $K$ be a point on $BC$,different from its midpoint.$D$ is on the extension of segment $AK,BD$ and $AC$,$CD$and$AB$intersect at $N,M$ respectively.prove that $A,B,D,C$ are concyclic.
Let $ABCD$ be a quadrilateral that can be inscribed in a circle. Denote by $P$ the intersection point of lines $AD$ and $BC$, and by $Q$ the intersection point of lines $AB$ and $DC$. Let $E$ be the fourth vertex of the parallelogram $ABCE$, and $F$ the intersection of lines $CE$ is $PQ$. Prove that the points $D,E, F$, and $Q$ lie on the same circle.
Let $ABC$ be an acute triangle with the circumcircle $k$ and incenter $I$. The perpendicular through $I$ in $CI$ intersects segment $[BC]$ in $U$ and $k$ in $V$. In particular $V$ and $A$ are on different sides of $BC$. The parallel line through $U$ to $AI$ intersects $AV$ in $X$.
Prove: If $XI$ and $AI$ are perpendicular to each other, then $XI$ intersects segment $[AC]$ in its midpoint $M$.
[i](Notation: $[\cdot]$ denotes the line segment.)[/i]
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 $h$ be a semicircle with diameter $AB$. The two circles $k_1$ and $k_2$, $k_1 \ne k_2$, touch the segment $AB$ at the points $C$ and $D$, respectively, and the semicircle $h$ fom the inside at the points $E$ and $F$, respectively. Prove that the four points $C$, $D$, $E$ and $F$ lie on a circle.
[i](Walther Janous)[/i]
Let $\triangle ABC$ be an acute angle triangle and $\omega$ be its circumcircle. Let $N$ be a point on arc $AC$ not containing $B$ and $S$ be a point on line $AB$. The line tangent to $\omega$ at $N$ intersects $BC$ at $T$, $NS$ intersects $\omega$ at $K$. Assume that $\angle NTC = \angle KSB$. Prove that $CK\parallel AN \parallel TS$.
Let $\omega_1$ and $\omega_2$ be two orthogonal circles, and let the center of $\omega_1$ be $O$. Diameter $AB$ of $\omega_1$ is selected so that $B$ lies strictly inside $\omega_2$. The two circles tangent to $\omega_2$, passing through $O$ and $A$, touch $\omega_2$ at $F$ and $G$. Prove that $FGOB$ is cyclic.
[i]Proposed by Eric Chen[/i]
We have $\vartriangle ABC$ with $I$ as its incenter. Let $D$ be the intersection of $AI$ and $BC$ and define $E, F$ in a similar way. Furthermore, let $Y = CI \cap DE, Z = BI \cap DF$. Prove that if $\angle BAC = 120^o$, then $E, F, Y,Z$ are concyclic.
[img]https://1.bp.blogspot.com/-5IFojUbPE3o/XnSKTlTISqI/AAAAAAAALd0/0OwKMl02KJgqPs-SDOlujdcWXM0cWJiegCK4BGAYYCw/s1600/imoc2017%2Bg5.png[/img]
Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$.
Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.
In convex quadrilateral $ABCD$, the diagonals $AC$ and $BD$ meet at the point $P$. We know that $\angle DAC = 90^o$ and $2 \angle ADB = \angle ACB$. If we have $ \angle DBC + 2 \angle ADC = 180^o$ prove that $2AP = BP$.
Proposed by Iman Maghsoudi
A trapezoid has side lengths as indicated in the figure (the sides with length $11$ and $36$ are parallel). Calculate the area of the trapezoid.[img]https://1.bp.blogspot.com/-5PKrqDG37X4/XzcJtCyUv8I/AAAAAAAAMY0/tB0FObJUJdcTlAJc4n6YNEaVIDfQ91-eQCLcBGAsYHQ/s0/1995%2BMohr%2Bp1.png[/img]
Denote by $ M$ midpoint of side $ BC$ in an isosceles triangle $ \triangle ABC$ with $ AC = AB$. Take a point $ X$ on a smaller arc $ \overarc{MA}$ of circumcircle of triangle $ \triangle ABM$. Denote by $ T$ point inside of angle $ BMA$ such that $ \angle TMX = 90$ and $ TX = BX$.
Prove that $ \angle MTB - \angle CTM$ does not depend on choice of $ X$.
[i]Author: Farzan Barekat, Canada[/i]
Let $ ABCD$ be a convex quadrilateral and let $ P$ and $ Q$ be points in $ ABCD$ such that $ PQDA$ and $ QPBC$ are cyclic quadrilaterals. Suppose that there exists a point $ E$ on the line segment $ PQ$ such that $ \angle PAE \equal{} \angle QDE$ and $ \angle PBE \equal{} \angle QCE$. Show that the quadrilateral $ ABCD$ is cyclic.
[i]Proposed by John Cuya, Peru[/i]
In triangle $ABC$ we have $AB = BC, \angle B = 20^o$. Point $M$ on $AC$ is such that $AM : MC = 1 : 2$, point $H$ is the projection of $C$ to $BM$. Find angle $AHB$.
(M. Yevdokimov)
In a concyclic quadrilateral $PQRS$,$\angle PSR=\frac{\pi}{2}$ , $H,K$ are perpendicular foot from $Q$ to sides $PR,RS$ , prove that $HK$ bisect segment$SQ$.
Let \( \triangle ABC \) be an isosceles triangle with \( AB = AC \). Let \( D \) be a point on \( AC \). Let \( L \) be a point inside the triangle such that \( \angle CLD = 90^\circ \) and
\[
CL \cdot BD = BL \cdot CD.
\]
Prove that the circumcenter of triangle \( \triangle BDL \) lies on line \( AB \).
Let $ABC$ be a triangle with $CA \neq CB$. Let $D$, $F$, and $G$ be the midpoints of the sides $AB$, $AC$, and $BC$ respectively. A circle $\Gamma$ passing through $C$ and tangent to $AB$ at $D$ meets the segments $AF$ and $BG$ at $H$ and $I$, respectively. The points $H'$ and $I'$ are symmetric to $H$ and $I$ about $F$ and $G$, respectively. The line $H'I'$ meets $CD$ and $FG$ at $Q$ and $M$, respectively. The line $CM$ meets $\Gamma$ again at $P$. Prove that $CQ = QP$.
[i]Proposed by El Salvador[/i]
Let $ABC$ be a triangle with circumcircle $\Omega$ and incentre $I$. Let the line passing through $I$ and perpendicular to $CI$ intersect the segment $BC$ and the arc $BC$ (not containing $A$) of $\Omega$ at points $U$ and $V$ , respectively. Let the line passing through $U$ and parallel to $AI$ intersect $AV$ at $X$, and let the line passing through $V$ and parallel to $AI$ intersect $AB$ at $Y$ . Let $W$ and $Z$ be the midpoints of $AX$ and $BC$, respectively. Prove that if the points $I, X,$ and $Y$ are collinear, then the points $I, W ,$ and $Z$ are also collinear.
[i]Proposed by David B. Rush, USA[/i]
Let $ABC$ be a triangle with incenter $I$. Two distinct points $P$ and $Q$ are chosen on the circumcircle of $ABC$ such that
\[ \angle API = \angle AQI = 45^\circ. \]
Lines $PQ$ and $BC$ meet at $S$. Let $H$ denote the foot of the altitude from $A$ to $BC$. Prove that $\angle AHI = \angle ISH$.
[i]Matsvei Zorka[/i]
In a triangle $ABC$, the external bisector of $\angle BAC$ intersects the ray $BC$ at $D$. The feet of the perpendiculars from $B$ and $C$ to line $AD$
are $E$ and $F$, respectively and the foot of the perpendicular from $D$ to $AC$ is $G$. Show that $\angle DGE + \angle DGF = 180^{\circ}$.
Let $\triangle ABC$ be an acute-angled triangle with $AB \not= AC$. Let $H$ be the orthocenter of triangle $ABC$, and let $M$ be the midpoint of the side $BC$. Let $D$ be a point on the side $AB$ and $E$ a point on the side $AC$ such that $AE=AD$ and the points $D$, $H$, $E$ are on the same line. Prove that the line $HM$ is perpendicular to the common chord of the circumscribed circles of triangle $\triangle ABC$ and triangle $\triangle ADE$.
Let the diagonals of cyclic quadrilateral $ABCD$ meet at point $P$. The line passing through $P$ and perpendicular to $PD$ meets $AD$ at point $D_1$, a point $A_1$ is defined similarly. Prove that the tangent at $P$ to the circumcircle of triangle $D_1PA_1$ is parallel to $BC$.
Let $I$ be the incenter of triangle $ABC$. The excircle with center $I_A$ touches the side $BC$ at point $A'$. The line $l$ passing through $I$ and perpendicular to $BI$ meets $I_AA'$ at point $K$ lying on the medial line parallel to $BC$. Prove that $\angle B \leq 60^\circ$.