Found problems: 847
Let $ABC$ be a triangle with $AB=AC$, and let $M$ be the midpoint of $BC$. Let $P$ be a point such that $PB<PC$ and $PA$ is parallel to $BC$. Let $X$ and $Y$ be points on the lines $PB$ and $PC$, respectively, so that $B$ lies on the segment $PX$, $C$ lies on the segment $PY$, and $\angle PXM=\angle PYM$. Prove that the quadrilateral $APXY$ is cyclic.
In triangle $ABC$, $X$ and $Y$ are the tangency points of incircle (with center $I$) with sides $AB$ and $AC$ respectively. A tangent line to the circumcircle of triangle $ABC$ (with center $O$) at point $A$, intersects the extension of $BC$ at $D$. If $D,X$ and $Y$ are collinear then prove that $D,I$ and $O$ are also collinear.
[i]proposed by Amirhossein Zabeti[/i]
A triangle $ABC$ is given. Let $M$ be the midpoint of the side $AC$ of the triangle and $Z$ the image of point $B$ along the line $BM$. The circle with center $M$ and radius $MB$ intersects the lines $BA$ and $BC$ at the points $E$ and $G$ respectively. Let $H$ be the point of intersection of $EG$ with the line $AC$, and $K$ the point of intersection of $HZ$ with the line $EB$. The perpendicular from point $K$ to the line $BH$ intersects the lines $BZ$ and $BH$ at the points $L$ and $N$, respectively.
If $P$ is the second point of intersection of the circumscribed circles of the triangles $KZL$ and $BLN$, prove that, the lines $BZ, KN$ and $HP$ intersect at a common point.
In the triangle $A,B,C$, let $D$ be the middle of $BC$ and $E$ the projection of $C$ on $AD$. Suppose $\angle ACE = \angle ABC$. Show that the triangle $ABC$ is isosceles or rectangle.
The circumferences $C_1$ and $C_2$ cut each other at different points $A$ and $K$. The common tangent to $C_1$ and $C_2$ nearer to $K$ touches $C_1$ at $B$ and $C_2$ at $C$. Let $P$ be the foot of the perpendicular from $B$ to $AC$, and let $Q$ be the foot of the perpendicular from $C$ to $AB$. If $E$ and $F$ are the symmetric points of $K$ with respect to the lines $PQ$ and $BC$, respectively, prove that $A, E$ and $F$ are collinear.
Initially, three non-collinear points, $A$, $B$, and $C$, are marked on the plane. You have a pencil and a double-edged ruler of width $1$. Using them, you may perform the following operations:
[list]
[*]Mark an arbitrary point in the plane.
[*]Mark an arbitrary point on an already drawn line.
[*]If two points $P_1$ and $P_2$ are marked, draw the line connecting $P_1$ and $P_2$.
[*]If two non-parallel lines $l_1$ and $l_2$ are drawn, mark the intersection of $l_1$ and $l_2$.
[*]If a line $l$ is drawn, draw a line parallel to $l$ that is at distance $1$ away from $l$ (note that two such lines may be drawn).
[/list]
Prove that it is possible to mark the orthocenter of $ABC$ using these operations.
The incircle of $ABC$ touches $BC$, $CA$, $AB$ at $K$, $L$, $M$ respectively. The line through $A$ parallel to $LK$ meets $MK$ at $P$, and the line through $A$ parallel to $MK$ meets $LK$ at $Q$. Show that the line $PQ$ bisects $AB$ and bisects $AC$.
Point $D$ lies inside triangle $ABC$ such that $\angle DAC = \angle DCA = 30^{\circ}$ and $\angle DBA = 60^{\circ}$. Point $E$ is the midpoint of segment $BC$. Point $F$ lies on segment $AC$ with $AF = 2FC$. Prove that $DE \perp EF$.
Given a cyclic quadrilateral $ABCD$, let the diagonals $AC$ and $BD$ meet at $E$ and the lines $AD$ and $BC$ meet at $F$. The midpoints of $AB$ and $CD$ are $G$ and $H$, respectively. Show that $EF$ is tangent at $E$ to the circle through the points $E$, $G$ and $H$.
[i]Proposed by David Monk, United Kingdom[/i]
Let $ABC$ be a scalene triangle and $\omega$ is your incircle. The sides $BC,CA$ and $AB$ are tangents to $\omega$ in $X,Y,Z$ respectively. Let $M$ be the midpoint of $BC$ and $D$ is the intersection point of $BC$ with the angle bisector of $\angle BAC$. Prove that $\angle BAX=\angle MAC$ if and only if $YZ$ passes by the midpoint of $AD$.
Let $ABC$ be a triangle with incenter $I$, and $A$-excenter $\Gamma$. Let $A_1,B_1,C_1$ be the points of tangency of $\Gamma$ with $BC,AC$ and $AB$, respectively. Suppose $IA_1, IB_1$ and $IC_1$ intersect $\Gamma$ for the second time at points $A_2,B_2,C_2$, respectively. $M$ is the midpoint of segment $AA_1$. If the intersection of $A_1B_1$ and $A_2B_2$ is $X$, and the intersection of $A_1C_1$ and $A_2C_2$ is $Y$, prove that $MX=MY$.
Let $ABC$ be a triangle. The incircle of triangle $ABC$ touches the side $BC$ at $A^{\prime}$, and the line $AA^{\prime}$ meets the incircle again at a point $P$. Let the lines $CP$ and $BP$ meet the incircle of triangle $ABC$ again at $N$ and $M$, respectively. Prove that the lines $AA^{\prime}$, $BN$ and $CM$ are concurrent.
Let $ABC$ be a triangle with $AB < AC$, incenter $I$, and $A$ excenter $I_{A}$. The incircle meets $BC$ at $D$. Define $E = AD\cap BI_{A}$, $F = AD\cap CI_{A}$. Show that the circumcircle of $\triangle AID$ and $\triangle I_{A}EF$ are tangent to each other
In $\triangle ABC$, point $M$ is the middle point of $AC$. $MD//AB$ and meet the tangent of $A$ to $\odot(ABC)$ at point $D$. Point $E$ is in $AD$ and point $A$ is the middle point of $DE$. $\{P\}=\odot(ABE)\cap AC,\{Q\}=\odot(ADP)\cap DM$. Prove that $\angle QCB=\angle BAC$.
[url=https://imgtu.com/i/4pZ7Zj][img]https://z3.ax1x.com/2021/09/12/4pZ7Zj.jpg[/img][/url]
Inscribed Triangle $ABC$ on circle $\odot O$. Bisector of $\angle ABC$ intersects $\odot O$ at $D$. Two lines $PB$ and $PC$ that are tangent to $\odot O$ intersect at $P$. $PD$ intersects $AC$ at $E$, $\odot O$ at $F$. $M$ is the midpoint of $BC$. Prove that $M,F,C,E$ are concyclic.
Let $ABC$ be a triangle such that $AB < AC$. Let $P$ lie on a line through $A$ parallel to line $BC$ such that $C$ and $P$ are on the same side of line $AB$. Let $M$ be the midpoint of segment $BC$. Define $D$ on segment $BC$ such that $\angle BAD = \angle CAM$, and define $T$ on the extension of ray $CB$ beyond $B$ so that $\angle BAT = \angle CAP$. Given that lines $PC$ and $AD$ intersect at $Q$, that lines $PD$ and $AB$ intersect at $R$, and that $S$ is the midpoint of segment $DT$, prove that if $A$,$P$,$Q$, and $R$ lie on a circle, then $Q$, $R$, and $S$ are collinear.
[i]David Rush[/i]
Let $ABC$ be an acute triangle with incenter $I$ and circumcircle $\Omega$. The line passing $I$ and perpendicular to $AI$ meets $AB, AC$ at $D, E$, respectively. $A$-excircle of $\triangle{ABC}$ meets $BC$ at $T$. $AT$ meets $\Omega$ at $P$. The line passing $P$ and parallel to $BC$ meets $\Omega$ at $Q$. The intersection of $QI$ and $AT$ is $K$. Prove that $Q,D,K,E$ are concyclic.
In triangle $ABC$ angle bisectors $AA_{1}$ and $CC_{1}$ intersect at $I$. Line through $B$ parallel to $AC$ intersects rays $AA_{1}$ and $CC_{1}$ at points $A_{2}$ and $C_{2}$ respectively. Let $O_{a}$ and $O_{c}$ be the circumcenters of triangles $AC_{1}C_{2}$ and $CA_{1}A_{2}$ respectively. Prove that $\angle{O_{a}BO_{c}} = \angle{AIC} $
Consider the family of nonisosceles triangles $ABC$ satisfying the property $AC^2 + BC^2 = 2 AB^2$. Points $M$ and $D$ lie on side $AB$ such that $AM = BM$ and $\angle ACD = \angle BCD$. Point $E$ is in the plane such that $D$ is the incenter of triangle $CEM$. Prove that exactly one of the ratios
\[ \frac{CE}{EM}, \quad \frac{EM}{MC}, \quad \frac{MC}{CE} \]
is constant.
Let a convex quadrilateral $ ABCD$ fixed such that $ AB \equal{} BC$, $ \angle ABC \equal{} 80, \angle CDA \equal{} 50$. Define $ E$ the midpoint of $ AC$; show that $ \angle CDE \equal{} \angle BDA$
[i](Paolo Leonetti)[/i]
Let $ABC$ be a triangle with $AB=AC$, and let $M$ be the midpoint of $BC$. Let $P$ be a point such that $PB<PC$ and $PA$ is parallel to $BC$. Let $X$ and $Y$ be points on the lines $PB$ and $PC$, respectively, so that $B$ lies on the segment $PX$, $C$ lies on the segment $PY$, and $\angle PXM=\angle PYM$. Prove that the quadrilateral $APXY$ is cyclic.
Let $ABC$ be a scalene triangle with circumcenter $O$, and let $D$ and $E$ be points inside angle $\measuredangle BAC$ such that $A$ lies on line $DE$, and $\angle ADB=\angle CBA$ and $\angle AEC=\angle BCA$. Let $M$ be the midpoint of $BC$ and $K$ be a point such that $OK$ is perpendicular to $AO$ and $\angle BAK=\angle MAC$. Finally, let $P$ be the intersection of the perpendicular bisectors of $BD$ and $CE$. Show that $KO=KP$.
[i]Proposed by Victor Domínguez[/i]
Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$, $AB$, respectively. Show that $PQ=QR$ if and only if the bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$.
Let $ ABC$ be a triangle with $ G$ as its centroid. Let $ P$ be a variable point on segment $ BC$. Points $ Q$ and $ R$ lie on sides $ AC$ and $ AB$ respectively, such that $ PQ \parallel AB$ and $ PR \parallel AC$. Prove that, as $ P$ varies along segment $ BC$, the circumcircle of triangle $ AQR$ passes through a fixed point $ X$ such that $ \angle BAG = \angle CAX$.
Two circles $\omega_1$ and $\omega_2$, of equal radius intersect at different points $X_1$ and $X_2$. Consider a circle $\omega$ externally tangent to $\omega_1$ at $T_1$ and internally tangent to $\omega_2$ at point $T_2$. Prove that lines $X_1T_1$ and $X_2T_2$ intersect at a point lying on $\omega$.