Found problems: 3585
Let $ABC$ be a triangle with $\angle{C} = 90^{\circ}$, and let $H$ be the foot of the altitude from $C$. A point $D$ is chosen inside the triangle $CBH$ so that $CH$ bisects $AD$. Let $P$ be the intersection point of the lines $BD$ and $CH$. Let $\omega$ be the semicircle with diameter $BD$ that meets the segment $CB$ at an interior point. A line through $P$ is tangent to $\omega$ at $Q$. Prove that the lines $CQ$ and $AD$ meet on $\omega$.
Let $ABC$ be an acute triangle, let $H$ and $O$ be its orthocentre and circumcentre, respectively,
and let $S$ and $T$ be the feet of the altitudes from $B$ to $AC$ and from $C$ to $AB$, respectively.
Let $M$ be the midpoint of the segment $ST$, and let $N$ be the midpoint of the segment $AH$. The line
through $O$, parallel to $BC$, crosses the sides $AC$ and $AB$ at $F$ and $G$, respectively. The line $NG$
meets the circle $BGO$ again at $K$, and the line $NF$ meets the circle $CFO$ again at $L$. Prove that
the triangles $BCM$ and $KLN$ are similar.
Given a cyclic $ABCD$ with $AB=AD$. Points $M$ and $N$ are marked on the sides $CD$ and $BC$, respectively, so that $DM+BN=MN$. Prove that the circumcenter of the triangle $AMN$ belongs to the segment $AC$.
N.Sedrakian
Circles $ w_{1}$ and $ w_{2}$ with centres $ O_{1}$ and $ O_{2}$ are externally tangent at point $ D$ and internally tangent to a circle $ w$ at points $ E$ and $ F$ respectively. Line $ t$ is the common tangent of $ w_{1}$ and $ w_{2}$ at $ D$. Let $ AB$ be the diameter of $ w$ perpendicular to $ t$, so that $ A, E, O_{1}$ are on the same side of $ t$. Prove that lines $ AO_{1}$, $ BO_{2}$, $ EF$ and $ t$ are concurrent.
Let $ABC$ be an acute triangle and $\Gamma$ its circumcircle. Let $l$ be the line tangent to $\Gamma$ at $A$. Let $D$ and $E$ be the intersections of the circumference with center $B$ and radius $AB$ with lines $l$ and $AC$, respectively. Prove the orthocenter of $ABC$ lies on line $DE$.
$ABCD$ is a cyclic quadrilateral and $M$ a point on the side $CD$ such that $ADM$ and $ABCM$ have the same area and the same perimeter. Show that two sides of $ABCD$ have the same length.
Let $ABCD$ be a convex quadrilateral with $\angle ABC = \angle ADC < 90^{\circ}$. The internal angle bisectors of $\angle ABC$ and $\angle ADC$ meet $AC$ at $E$ and $F$ respectively, and meet each other at point $P$. Let $M$ be the midpoint of $AC$ and let $\omega$ be the circumcircle of triangle $BPD$. Segments $BM$ and $DM$ intersect $\omega$ again at $X$ and $Y$ respectively. Denote by $Q$ the intersection point of lines $XE$ and $YF$. Prove that $PQ \perp AC$.
Let $ABCD$ be a cyclic quadrilateral with opposite sides not parallel. Let $X$ and $Y$ be the intersections of $AB,CD$ and $AD,BC$ respectively. Let the angle bisector of $\angle AXD$ intersect $AD,BC$ at $E,F$ respectively, and let the angle bisectors of $\angle AYB$ intersect $AB,CD$ at $G,H$ respectively. Prove that $EFGH$ is a parallelogram.
Let $ABC$ be a triangle. $M$ the midpoint of $AB$ and $L$ the midpoint of $BC$. We denote by $G$ the intersection of $AL$ with $CM$ and we take $E$ a point such that $G$ is the midpoint of the segment $AE$. Prove that the quadrilateral $MCEB$ is cyclic if and only if $MB = BG$.
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]
In triangle $ABC, \angle A$ is smaller than $\angle C$. Point $D$ lies on the (extended) line $BC$ (with $B$ between $C$ and $D$) such that $|BD| = |AB|$. Point $E$ lies on the bisector of $\angle ABC$ such that $\angle BAE = \angle ACB$. Line segment $BE$ intersects line segment $AC$ in point $F$. Point $G$ lies on line segment $AD$ such that $EG$ and $BC$ are parallel. Prove that $|AG| =|BF|$.
[asy]
unitsize (1.5 cm);
real angleindegrees(pair A, pair B, pair C) {
real a, b, c;
a = abs(B - C);
b = abs(C - A);
c = abs(A - B);
return(aCos((a^2 + c^2 - b^2)/(2*a*c)));
};
pair A, B, C, D, E, F, G;
B = (0,0);
A = 2*dir(190);
D = 2*dir(310);
C = 1.5*dir(310 - 180);
E = extension(B, incenter(A,B,C), A, rotate(angleindegrees(A,C,B),A)*(B));
F = extension(B,E,A,C);
G = extension(E, E + D - B, A, D);
filldraw(anglemark(A,C,B,8),gray(0.8));
filldraw(anglemark(B,A,E,8),gray(0.8));
draw(C--A--B);
draw(E--A--D);
draw(interp(C,D,-0.1)--interp(C,D,1.1));
draw(interp(E,B,-0.2)--interp(E,B,1.2));
draw(E--G);
dot("$A$", A, SW);
dot("$B$", B, NE);
dot("$C$", C, NE);
dot("$D$", D, NE);
dot("$E$", E, N);
dot("$F$", F, N);
dot("$G$", G, SW);
[/asy]
Two quarter-circles touch as shown. Find the angle $x$.
[img]https://cdn.artofproblemsolving.com/attachments/b/4/e70d5d69e46d6d40368f143cb83cf10b7d6d98.png[/img]
In a convex quadrilateral $ABCD$, $\angle ABD=30^\circ$, $\angle BCA=75^\circ$, $\angle ACD=25^\circ$ and $CD=CB$. Extend $CB$ to meet the circumcircle of triangle $DAC$ at $E$. Prove that $CE=BD$.
[i]Proposed by BJ Venkatachala[/i]
Let $ABC$ be a triangle with incenter $I$. Suppose the reflection of $AB$ across $CI$ and the reflection of $AC$ across $BI$ intersect at a point $X$. Prove that $XI$ is perpendicular to $BC$.
A point $D$ is chosen on the side $AC$ of a triangle $ABC$ with $\angle C < \angle A < 90^\circ$ in such a way that $BD=BA$. The incircle of $ABC$ is tangent to $AB$ and $AC$ at points $K$ and $L$, respectively. Let $J$ be the incenter of triangle $BCD$. Prove that the line $KL$ intersects the line segment $AJ$ at its midpoint.
Let $ABC$ be an acute triangle orthocenter angle $H$. Let $\omega_1$ be the circle tangent to $BC$ at $B$ and passing through $H$ and $\omega_2$ the circle tangent to $BC$ at $C$ and passing through through $H$. A line $\ell$ passing through $H$ intersects the circles $\omega_1$ and $\omega_2$ at points $D$ and $E$, respectively (with $D$ and $E$ other than $H$). Lines $BD$ and $CE$ intersect at $F$, the lines $\ell$ and $AF$ intersect at $X$ and the circles $\omega_1$ and $\omega_2$ intersect at the points $P$ and $H$. Prove that the points $A, H, P$ and $X$ are still on the same circle.
Cyclic quadrilateral $ABCD$ with circumcircle $\omega$ is given. Let $E$ be a fixed point on segment $AC$. $M$ is an arbitrary point on $\omega$, lines $AM$ and $BD$ meet at a point $P$. $EP$ meets $AB$ and $AD$ at points $R$ and $Q$, respectively, $S$ is the intersection of $BQ,DR$ and lines $MS$ and $AC$ meet at a point $T$. Prove that as $M$ varies the circumcircle of triangle $\bigtriangleup CMT$ passes through a fixed point other than $C$.
[i]Proposed by Chunlai Jin - China[/i]
Let $ABCD$ be a circumscribed quadrilateral. Let $g$ be a line through $A$ which meets the segment $BC$ in $M$ and the line $CD$ in $N$. Denote by $I_1$, $I_2$ and $I_3$ the incenters of $\triangle ABM$, $\triangle MNC$ and $\triangle NDA$, respectively. Prove that the orthocenter of $\triangle I_1I_2I_3$ lies on $g$.
[i]Proposed by Nikolay Beluhov, Bulgaria[/i]
Triangle $BCF$ has a right angle at $B$. Let $A$ be the point on line $CF$ such that $FA=FB$ and $F$ lies between $A$ and $C$. Point $D$ is chosen so that $DA=DC$ and $AC$ is the bisector of $\angle{DAB}$. Point $E$ is chosen so that $EA=ED$ and $AD$ is the bisector of $\angle{EAC}$. Let $M$ be the midpoint of $CF$. Let $X$ be the point such that $AMXE$ is a parallelogram. Prove that $BD,FX$ and $ME$ are concurrent.
Let $C$ and $D$ be two points on the semicricle with diameter $AB$ such that $B$ and $C$ are on distinct sides of the line $AD$. Denote by $M$, $N$ and $P$ the midpoints of $AC$, $BD$ and $CD$ respectively. Let $O_A$ and $O_B$ the circumcentres of the triangles $ACP$ and $BDP$. Show that the lines $O_AO_B$ and $MN$ are parallel.
Let $ABC$ be a triangle with circumcenter $O$. Point $X$ is the intersection of the parallel line from $O$ to $AB$ with the perpendicular line to $AC$ from $C$. Let $Y$ be the point where the external bisector of $\angle BXC$ intersects with $AC$. Let $K$ be the projection of $X$ onto $BY$. Prove that the lines $AK, XO, BC$ have a common point.
Let $ABC$ be a scalene triangle with orthocenter $H$ and circumcenter $O$. Let $P$ be the midpoint of $\overline{AH}$ and let $T$ be on line $BC$ with $\angle TAO=90^{\circ}$. Let $X$ be the foot of the altitude from $O$ onto line $PT$. Prove that the midpoint of $\overline{PX}$ lies on the nine-point circle* of $\triangle ABC$.
*The nine-point circle of $\triangle ABC$ is the unique circle passing through the following nine points: the midpoint of the sides, the feet of the altitudes, and the midpoints of $\overline{AH}$, $\overline{BH}$, and $\overline{CH}$.
[i]Proposed by Zack Chroman[/i]
In triangle $ABC$ let $N, M, P$ be the midpoints of the sides $BC, CA, AB$ and $G$ be the centroid of this triangle. Let the circle circumscribed to $BGP$ intersect the line $MP$ in point $K$, $P \neq K$, and the circle circumscribed to $CGN$ intersect the line $MN$ in point $L$, $N \neq L$. Prove that $ \angle BAK = \angle CAL $.
Let $\triangle ABC$ be an acute triangle. The line through $A$ perpendicular to $BC$ intersects $BC$ at $D$.
Let $E$ be the midpoint of $AD$ and $\omega$ the the circle with center $E$ and radius equal to $AE$. The line
$BE$ intersects $\omega$ at a point $X$ such that $X$ and $B$ are not on the same side of $AD$ and the line $CE$
intersects $\omega$ at a point $Y$ such that $C$ and $Y$ are not on the same side of $AD$. If both of the intersection
points of the circumcircles of $\triangle BDX$ and $\triangle CDY$ lie on the line $AD$, prove that $AB = AC$.
Given an acute-angled triangle $ABC$ in which $|AB| <|AC|$. Point $D$ is the midpoint of the shorter arc $BC$ of its circumcircle. The point $I$ is the center of its incircle, and the point $J$ is symmetric point of $I$ wrt line $BC$. The line $DJ$ intersects the circumcircle of the triangle $ABC$ at the point $E$ belonging to the arc $AB$. Prove that $|AI |= |IE|$.