Given a triangle $ABC$, the line passing through the vertex $A$ symmetric to the median $AM$ wrt the line containing the bisector of the angle $\angle BAC$ intersects the circle circumscribed around the triangle $ABC$ at points $A$ and $K$. Let $L$ be the midpoint of the segment $AK$. Prove that $\angle BLC=2\angle BAC$.

Let $ABC$ be a triangle with circumcircle $k$. The tangents at $A$ and $C$ intersect at $T$. The circumcircle of triangle $ABT$ intersects the line $CT$ at $X$ and $Y$ is the midpoint of $CX$. Prove that the lines $AX$ and $BY$ intersect on $k$.

The altitudes $AA_1, BB_1$ and $CC_1$ were drawn in the triangle $ABC$. Point $K$ is a projection of point $B$ on $A_1C_1$. Prove that the symmmedian $\vartriangle ABC$ from the vertex $B$ divides the segment $B_1K$ in half. (Anton Trygub)

Let a point $D$ lie on the median $AM$ of a triangle $ABC$. The tangents to the circumcircle of triangle $BDC$ at points $B$ and $C$ meet at point $K$. Prove that $DD'$ is parallel to $AK$, where $D'$ is isogonally conjugated to $D$ with respect to $ABC$.

Given triangle $ABC$ with the inner - bisector $AD$. The line passes through $D$ and perpendicular to $BC$ intersects the outer - bisector of $\angle BAC$ at $I$. Circle $(I,ID)$ intersects $CA$, $AB$ at $E$, $F$, reps. The symmedian line of $\triangle AEF$ intersects the circle $(AEF)$ at $X$. Prove that the circles $(BXC)$ and $(AEF)$ are tangent. [Hide=Diagram] [asy]import graph; size(7.04cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 7.02, xmax = 14.06, ymin = -1.54, ymax = 4.08; /* image dimensions */ /* draw figures */ draw((8.62,3.12)--(7.58,-0.38)); draw((7.58,-0.38)--(13.68,-0.38)); draw((13.68,-0.38)--(8.62,3.12)); draw((8.62,3.12)--(9.85183961338573,3.5535510951732316)); draw((9.85183961338573,3.5535510951732316)--(9.851839613385732,-0.38)); draw((9.851839613385732,-0.38)--(8.62,3.12)); draw(circle((10.012708209519483,1.129702986881574), 2.4291805937992947)); draw((8.62,3.12)--(9.470868507287285,-1.238276762688951), red); draw(shift((9.85183961338573,3.553551095173232))*xscale(3.9335510951732324)*yscale(3.9335510951732324)*arc((0,0),1,237.85842690125605,309.7357733435313), linetype("4 4")); draw(shift((10.63,3.8274278922585725))*xscale(5.196628663716066)*yscale(5.196628663716066)*arc((0,0),1,234.06132677886183,305.9386732211382), blue); /* dots and labels */ dot((8.62,3.12),linewidth(3.pt) + dotstyle); label("$A$", (8.48,3.24), NE * labelscalefactor); dot((7.58,-0.38),linewidth(3.pt) + dotstyle); label("$B$", (7.3,-0.58), NE * labelscalefactor); dot((13.68,-0.38),linewidth(3.pt) + dotstyle); label("$C$", (13.76,-0.26), NE * labelscalefactor); dot((9.851839613385732,-0.38),linewidth(3.pt) + dotstyle); label("$D$", (9.94,-0.26), NE * labelscalefactor); dot((9.85183961338573,3.5535510951732316),linewidth(3.pt) + dotstyle); label("$I$", (9.94,3.68), NE * labelscalefactor); dot((7.759138898806625,0.22287129406075654),linewidth(3.pt) + dotstyle); label("$F$", (7.46,0.16), NE * labelscalefactor); dot((12.36635458796946,0.5286480122740898),linewidth(3.pt) + dotstyle); label("$E$", (12.44,0.64), NE * labelscalefactor); dot((9.470868507287285,-1.238276762688951),linewidth(3.pt) + dotstyle); label("$X$", (9.56,-1.12), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy] [/Hide]

Let $ABC$ be a triangle, and let $M$ be a point on the side $(AC)$ .The line through $M$ and parallel to $BC$ crosses $AB$ at $N$. Segments $BM$ and $CN$ cross at $P$, and the circles $BNP$ and $CMP$ cross again at $Q$. Show that angles $BAP$ and $CAQ$ are equal.

To two circles $r_1$ and $r_2$, intersecting at points $A$ and $B$, their common tangent $CD$ is drawn ($C$ and $D$ are tangency points, respectively, point $B$ is closer to line $CB$ than $A$). Line passing through $A$ , intersects $r_1$ and $r_2$ for second time at points $K$ and $L$, respectively ($A$ lies between $K$ and $L$). Lines $KC$ and $LD$ intersect at point $P$. Prove that $PB$ is the symmedian of triangle $KPL$. (Yu. Blinkov)

Let $ABC$ be a triangle inscribed in circle $(O)$, with its altitudes $BH_b, CH_c$ intersect at orthocenter $H$ ($H_b \in AC$, $H_c \in AB$). $H_bH_c$ meets $BC$ at $P$. Let $N$ be the midpoint of $AH, L$ be the orthogonal projection of $O$ on the symmedian with respect to angle $A$ of triangle $ABC$. Prove that $\angle NLP = 90^o$.

Let $ABC$ be a scalene acute-angled triangle and $D$ be the point on its circumcircle such that $AD$ is a symmedian of triangle $ABC$. Let $E$ be the reflection of $D$ about $BC$, $C_0$ the reflection of $E$ about $AB$ and $B_0$ the reflection of $E$ about $AC$. Prove that the lines $AD$, $BB_0$ and $CC_0$ are concurrent if and only if $\angle BAC = 60^{\circ}.$

Let $ABC$ be an acute-angled triangle with $AB<AC<BC$ and $c(O,R)$ the circumscribed circle. Let $D, E$ be points in the small arcs $AC, AB$ respectively. Let $K$ be the intersection point of $BD,CE$ and $N$ the second common point of the circumscribed circles of the triangles $BKE$ and $CKD$. Prove that $A, K, N$ are collinear if and only if $K$ belongs to the symmedian of $ABC$ passing from $A$.

Let $k$ be a circle with centre $O$. Let $AB$ be a chord of this circle with midpoint $M\neq O$. The tangents of $k$ at the points $A$ and $B$ intersect at $T$. A line goes through $T$ and intersects $k$ in $C$ and $D$ with $CT < DT$ and $BC = BM$. Prove that the circumcentre of the triangle $ADM$ is the reflection of $O$ across the line $AD$.

Let $ABC$ be a triangle and let $O$ be its circumcentre. The internal and external bisectrices of the angle $BAC$ meet the line $BC$ at points $D$ and $E$, respectively. Let further $M$ and $L$ respectively denote the midpoints of the segments $BC$ and $DE$. The circles $ABC$ and $ALO$ meet again at point $N$. Show that the angles $BAN$ and $CAM$ are equal.

Consider an acute-angled triangle $ABC$ with $AB<AC$ and let $\omega$ be its circumscribed circle. Let $t_B$ and $t_C$ be the tangents to the circle $\omega$ at points $B$ and $C$, respectively, and let $L$ be their intersection. The straight line passing through the point $B$ and parallel to $AC$ intersects $t_C$ in point $D$. The straight line passing through the point $C$ and parallel to $AB$ intersects $t_B$ in point $E$. The circumcircle of the triangle $BDC$ intersects $AC$ in $T$, where $T$ is located between $A$ and $C$. The circumcircle of the triangle $BEC$ intersects the line $AB$ (or its extension) in $S$, where $B$ is located between $S$ and $A$. Prove that $ST$, $AL$, and $BC$ are concurrent. $\text{Vangelis Psychas and Silouanos Brazitikos}$

Let $ABC$ be an isosceles triangle, $AB=AC$ inscribed in a circle $\omega$. The $B$-symmedian intersects $\omega$ again at $D$. The circle through $C,D$ and tangent to $BC$ and the circle through $A,D$ and tangent to $CD$ intersect at points $D,X$. The incenter of $ABC$ is denoted $I$. Prove that $B,C,I,X$ are concyclic.

Let a triangle $ABC$ be given. Consider the circle touching its circumcircle at $A$ and touching externally its incircle at some point $A_1$. Points $B_1$ and $C_1$ are defined similarly. a) Prove that lines $AA_1, BB_1$ and $CC1$ concur. b) Let $A_2$ be the touching point of the incircle with $BC$. Prove that lines $AA_1$ and $AA_2$ are symmetric about the bisector of angle $\angle A$.

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 $ABC$ be a triangle and let $X$ and $Y$ be points on the $A$-symmedian such that $AX = XB$ and $AY = YC$. Let $BX$ and $CY$ meet at $Z$. Let the $Z$-excircle of triangle $XYZ$ touch $ZX$ and $ZY$ at $E$ and $F$. Show that $A$, $E$, $F$ are collinear.

Let $AD$ be the inner angle bisector of the triangle $ABC$. The perpendicular on the side $BC$ at the point $D$ intersects the outer bisector of $\angle CAB$ at point $I$. The circle with center $I$ and radius $ID$ intersects the sides $AB$ and $AC$ at points $F$ and $E$ respectively. $A$-symmedian of $\Delta AFE$ intersects the circumcircle of $\Delta AFE$ again at point $X$. Prove that the circumcircles of $\Delta AFE$ and $\Delta BXC$ are tangent.

In a triangle $\triangle ABC$ with$\angle B>\angle C$, the altitude, the angle bisector, and the median from $A$ intersect $BC$ at $H, L$ and $D$, respectively. Show that $\angle HAL=\angle DAL$ if and only if $\angle BAC=90^{\circ}$.

Consider an acute-angled triangle $ABC$ with $AB<AC$ and let $\omega$ be its circumscribed circle. Let $t_B$ and $t_C$ be the tangents to the circle $\omega$ at points $B$ and $C$, respectively, and let $L$ be their intersection. The straight line passing through the point $B$ and parallel to $AC$ intersects $t_C$ in point $D$. The straight line passing through the point $C$ and parallel to $AB$ intersects $t_B$ in point $E$. The circumcircle of the triangle $BDC$ intersects $AC$ in $T$, where $T$ is located between $A$ and $C$. The circumcircle of the triangle $BEC$ intersects the line $AB$ (or its extension) in $S$, where $B$ is located between $S$ and $A$. Prove that $ST$, $AL$, and $BC$ are concurrent. $\text{Vangelis Psychas and Silouanos Brazitikos}$

Restore a triangle $ABC$ by vertex $B$, the centroid and the common point of the symmedian from $B$ with the circumcircle.

Let $\overline{AB}$ be a chord of a circle $\omega$, and let $P$ be a point on the chord $\overline{AB}$. Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$. Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$. Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$. Line $PQ$ intersects $\omega$ at $X$ and $Y$. Assume that $AP=5$, $PB=3$, $XY=11$, and $PQ^2 = \tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Let $ABC$ be an acute triangle with altitudes $AA'$ and $BB'$ and orthocenter $H$. Let $C_0$ be the midpoint of the segment $AB$. Let $g$ be the line symmetric to the line $CC_0$ with respect to the angular bisector of $\angle ACB$. Let $h$ be the line symmetric to the line $HC_0$ with respect to the angular bisector of $\angle AHB$. Show that the lines $g$ and $h$ intersect on the line $A'B'$.

Let $ABC$ be an obtuse-angled triangle with $ \angle ABC>90^{\circ}$, and let $(c)$ be its circumcircle. The internal angle bisector of $\angle BAC$ meets again the circle $(c)$ at the point $E$, and the line $BC$ at the point $D$. The circle of diameter $DE$ meets the circle $(c)$ at the point $H$. If the line $HE$ meets the line $BC$ at the point $K$, prove that: (a) the points $K, H, D$ and $A$ are concyclic (b) the line $AH$ passes through the point of intersection of the tangents to the circle $(c)$ at the points $B$ and $C$.

It is known about the triangle $ABC$ that $3 BC = CA + AB$. Let the $A$-symmedian of triangle $ABC$ intersect the circumcircle of triangle $ABC$ at point $D$. Prove that $\frac{1}{BD}+ \frac{1}{CD}= \frac{6}{AD}$. (Ercole Suppa, Italy)