Found problems: 3585
Given an acute triangle $ PBC$ with $ PB\neq PC.$ Points $ A,D$ lie on $ PB,PC,$ respectively. $ AC$ intersects $ BD$ at point $ O.$ Let $ E,F$ be the feet of perpendiculars from $ O$ to $ AB,CD,$ respectively. Denote by $ M,N$ the midpoints of $ BC,AD.$
$ (1)$: If four points $ A,B,C,D$ lie on one circle, then $ EM\cdot FN \equal{} EN\cdot FM.$
$ (2)$: Determine whether the converse of $ (1)$ is true or not, justify your answer.
Let $ABDC$ be a cyclic quadrilateral inscribed in a circle $\Omega$. $AD$ meets $BC$ at $P$, and $\Omega$ meets lines passing $A$ and parallel to $DB$, $DC$ at $E$, $F$, respectively. $X$ is a point on $\Omega$ such that $PA=PX$. Prove that the lines $BE$, $CF$, and $DX$ are concurrent.
Quadrangle $ABCD$ is inscribed in a circle with diameter $AC$. Points $K$ and $M$ are projections of vertices $A$ and $C$, respectively, onto line $BD$. A line parallel to $BC$ is drawn through point $K$ and intersecting $AC$ at point $P$. Prove that angle $KPM$ is a right angle.
The incircle of triangle $ABC$ touches the sides in $X$, $Y$ and $Z$. In triangle $XYZ$ the feet of the altitude from $X$ and $Y$ are $X'$ and $Y'$, respectively. Let line $X'Y'$ intersect the circumcircle of triangle $ABC$ at $P$ and $Q$. Prove that points $X$, $Y$, $P$ and $Q$ are concyclic.
Proposed by [i]László Simon[/i], Budapest
In the triangle $ABC$ the incircle is tangent to the sides $AB$ and $AC$ at $D$ and $E$ respectively. The line $DE$ intersects the circumcircle at $P$ and $Q$, with $P$ in the small arc $AB$ and $Q$ in the small arc $AC$. If $P$ is the midpoint of the arc $AB$, find the angle A and the ratio $\frac{PQ}{BC}$.
Let $ABC$ be a nondegenerate acute triangle with circumcircle $\omega$ and let its incircle $\gamma$ touch $AB, AC, BC$ at $X, Y, Z$ respectively. Let $XY$ hit arcs $AB, AC$ of $\omega$ at $M, N$ respectively, and let $P \neq X, Q \neq Y$ be the points on $\gamma$ such that $MP=MX, NQ=NY$. If $I$ is the center of $\gamma$, prove that $P, I, Q$ are collinear if and only if $\angle BAC=90^\circ$.
[i]Proposed by David Stoner[/i]
In an acute triangle $ABC$, $AB \neq AC$, $O$ is its circumcenter. $K$ is the reflection of $B$ over $AC$ and $L$ is the reflection of $C$ over $AB$. $X$ is a point within $ABC$ such that $AX \perp BC, XK=XL$. Points $Y, Z$ are on $\overline{BK}, \overline{CL}$ respectively, satisfying $XY \perp CK, XZ \perp BL$.
Proof that $B, C, Y, O, Z$ lie on a circle.
In a cyclic quadrilateral $ABCD$ whose largest interior angle is $D$, lines $BC$ and $AD$ intersect at point $E$, while lines $AB$ and $CD$ intersect at point $F$. A point $P$ is taken in the interior of quadrilateral $ABCD$ for which $\angle EPD=\angle FPD=\angle BAD$. $O$ is the circumcenter of quadrilateral $ABCD$. Line $FO$ intersects the lines $AD$, $EP$, $BC$ at $X$, $Q$, $Y$, respectively. If $\angle DQX = \angle CQY$, show that $\angle AEB=90^\circ$.
Let $ABC$ be a triangle with circumcircle $\omega$ and let $\Omega_A$ be the $A$-excircle. Let $X$ and $Y$ be the intersection points of $\omega$ and $\Omega_A$. Let $P$ and $Q$ be the projections of $A$ onto the tangent lines to $\Omega_A$ at $X$ and $Y$ respectively. The tangent line at $P$ to the circumcircle of the triangle $APX$ intersects the tangent line at $Q$ to the circumcircle of the triangle $AQY$ at a point $R$. Prove that $\overline{AR} \perp \overline{BC}$.
let the incircle of a triangle ABC touch BC,AC,AB at A1,B1,C1 respectively. M and N are the midpoints of AB1 and AC1 respectively. MN meets A1C1 at T . draw two tangents TP and TQ through T to incircle. PQ meets MN at L and B1C1 meets PQ at K . assume I is the center of the incircle .
prove IK is parallel to AL
Let $ ABC $ be an acute triangle. Denote by $ D $ the foot of the perpendicular line drawn from the point $ A $ to the side $ BC $, by $M$ the midpoint of $ BC $, and by $ H $ the orthocenter of $ ABC $. Let $ E $ be the point of intersection of the circumcircle $ \Gamma $ of the triangle $ ABC $ and the half line $ MH $, and $ F $ be the point of intersection (other than $E$) of the line $ ED $ and the circle $ \Gamma $. Prove that $ \tfrac{BF}{CF} = \tfrac{AB}{AC} $ must hold.
(Here we denote $XY$ the length of the line segment $XY$.)
Let $ABCD$ be a quadrilateral $\angle B = \angle D$ and $AD = CD$. The incircle of triangle $ABC$ touches the sides $BC$ and $AB$ at points $E$ and $F$ respectively. Prove that the midpoints of segments $AC, BD, AE,$ and $CF$ are concyclic.
Circle $\omega$ is inscribed in quadrilateral $ABCD$ and is tangent to segments $BC, AD$ at $E,F$ , respectively.$DE$ intersects $\omega$ for the second time at $X$. if the circumcircle of triangle $DFX$ is tangent to lines $AB$ and $CD$ , prove that quadrilateral $AFXC$ is cyclic.
The incircle of a triangle $ABC$ touches the sides $AB,BC,CA$ at points $D,E,F$ respectively. The line through $A$ parallel to $DF$ meets the line through $C$ parallel to $EF$ at $G$.
$(a)$ Prove that the quadrilateral $AICG$ is cyclic.
$(b)$ Prove that the points $B,I,G$ are collinear.
Let $ABC$ be a triangle, and let $T$ be a point on the extension of $AB$ beyond $B$, and $U$ a point on the extension of $AC$ beyond $C$, such that $BT = CU$. Moreover, let $R$ and $S$ be points on the extensions of $AB$ and $AC$ beyond $A$ such that $AS = AT$ and $AR = AU$. Prove that $R$, $S$, $T$, $U$ lie on a circle whose centre lies on the circumcircle of $ABC$.
Let $I$ be the incenter of a triangle $ABC$. The points $X, Y$ lie on the prolongations of the lines $IB, IC$ after $I$ so that $\angle IAX=\angle IBA$ and $\angle IAY=\angle ICA$. Show that the line through the midpoints of $IA$ and $XY$ passes through the circumcenter of $ABC$.
In a triangle $ABC$, let $I$ denote its incenter. Points $D, E, F$ are chosen on the segments $BC, CA, AB$, respectively, such that $BD + BF = AC$ and $CD + CE = AB$. The circumcircles of triangles $AEF, BFD, CDE$ intersect lines $AI, BI, CI$, respectively, at points $K, L, M$ (different from $A, B, C$), respectively. Prove that $K, L, M, I$ are concyclic.
Let the sides $AD$ and $BC$ of the quadrilateral $ABCD$ (such that $AB$ is not parallel to $CD$) intersect at point $P$. Points $O_1$ and $O_2$ are circumcenters and points $H_1$ and $H_2$ are orthocenters of triangles $ABP$ and $CDP$, respectively. Denote the midpoints of segments $O_1H_1$ and $O_2H_2$ by $E_1$ and $E_2$, respectively. Prove that the perpendicular from $E_1$ on $CD$, the perpendicular from $E_2$ on $AB$ and the lines $H_1H_2$ are concurrent.
[i]Proposed by Eugene Bilopitov, Ukraine[/i]
The triangle $ABC$ is inscribed in a circle $w_1$. Inscribed in a triangle circle touchs the sides $BC$ in a point $N$. $w_2$ — the circle inscribed in a segment $BAC$ circle of $w_1$, and passing through a point $N$. Let points $O$ and $J$ — the centers of circles $w_2$ and an extra inscribed circle (touching side $BC$) respectively. Prove, that lines $AO$ and $JN$ are parallel.
Quadrilateral $XABY$ is inscribed in the semicircle $\omega$ with diameter $XY$. Segments $AY$ and $BX$ meet at $P$. Point $Z$ is the foot of the perpendicular from $P$ to line $XY$. Point $C$ lies on $\omega$ such that line $XC$ is perpendicular to line $AZ$. Let $Q$ be the intersection of segments $AY$ and $XC$. Prove that \[\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}.\]
Given trapezoid $ABCD$ with parallel sides $AB$ and $CD$, let $E$ be a point on line $BC$ outside segment $BC$, such that segment $AE$ intersects segment $CD$. Assume that there exists a point $F$ inside segment $AD$ such that $\angle EAD=\angle CBF$. Denote by $I$ the point of intersection of $CD$ and $EF$, and by $J$ the point of intersection of $AB$ and $EF$. Let $K$ be the midpoint of segment $EF$, and assume that $K$ is different from $I$ and $J$.
Prove that $K$ belongs to the circumcircle of $\triangle ABI$ if and only if $K$ belongs to the circumcircle of $\triangle CDJ$.
Let $ABC$ be an acute triangle. Denote by $H$ and $M$ the orthocenter of $ABC$ and the midpoint
of side $BC,$ respectively. Let $Y$ be a point on $AC$ such that $YH$ is perpendicular to $MH$ and let $Q$ be a point
on $BH$ such that $QA$ is perpendicular to $AM.$ Let $J$ be the second point of intersection of $MQ$ and the circle
with diameter $MY.$ Prove that $HJ$ is perpendicular to $AM.$
(Steve Dinh)
Let $ABC$ an acute triangle and $\Gamma$ its circumcircle. The bisector of $BAC$ intersects $\Gamma$ at $M\neq A$. A line $r$ parallel to $BC$ intersects $AC$ at $X$ and $AB$ at $Y$. Also, $MX$ and $MY$ intersect $\Gamma$ again at $S$ and $T$, respectively.
If $XY$ and $ST$ intersect at $P$, prove that $PA$ is tangent to $\Gamma$.
Let $ABCD$ be a cyclic quadrilateral, so that $|AB| + |CD| = |BC|$. Show that the intersection of the bisector of $\angle DAB$ and $\angle CDA$ lies on the side $BC$.
Let $ABCD$ be a quadrilateral inscribed in a circle $k$. $AC$ and $BD$ meet at $E$. The rays $\overrightarrow{CB}, \overrightarrow{DA}$ meet at $F$.
Prove that the line through the incenters of $\triangle ABE\,,\, \triangle ABF$ and the line through the incenters of $\triangle CDE\,,\, \triangle CDF$ meet at a point lying on the circle $k$.
[i]Proposed by N. Beluhov[/i]