If $AG_a,BG_b$, and $CG_c$ are symmedians in $\Delta ABC$ ($G_a\in BC,G_b\in AC,G_c\in AB$), is it possible for $\Delta G_a G_b G_c$ to be equilateral when $\Delta ABC$ is not equilateral?

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. $\text{Vangelis Psychas and Silouanos Brazitikos}$

Triangle $ABC$ has circumcircle $(O,R)$, and orthocenter $H$. The symmedians through $A,B,C$ meet the perpendicular bisectors of $BC,CA,AB$ at $D,E, F$ respectively. Let $M,N, P$ be the perpendicular projections of H on the line $OD,OE,OF.$ Prove that $$\frac{OH^2}{R^2} =\frac{\overline{OM}}{\overline{OD}}+\frac{\overline{ON}}{\overline{OE}} +\frac{\overline{OP}}{\overline{OF}}$$ Đỗ Thanh Sơn

On a triangle $ABC$, points $D$, $E$, $F$ are given on the segments $BC$, $AC$, $AB$ respectively such that $DE \parallel AB$, $DF \parallel AC$ and $\frac{BD}{DC}=\frac{AB^2}{AC^2}$ holds. Let the circumcircle of $AEF$ meet $AD$ at $R$ and the line that is tangent to the circumcircle of $ABC$ at $A$ at $S$ again. Let the line $EF$ intersect $BC$ at $L$ and $SR$ at $T$. Prove that $SR$ bisects $AB$ if and only if $BS$ bisects $TL$.

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)

For $\triangle ABC$ and its circumcircle $\omega$, draw the tangents at $B,C$ to $\omega$ meeting at $D$. Let the line $AD$ meet the circle with center $D$ and radius $DB$ at $E$ inside $\triangle ABC$. Let $F$ be the point on the extension of $EB$ and $G$ be the point on the segment $EC$ such that $\angle AFB=\angle AGE=\angle A$. Prove that the tangent at $A$ to the circumcircle of $\triangle AFG$ is parallel to $BC$. [i]Proposed by 61plus[/i]

Triangle $ABC$ is inscribed in circle $(O)$. $ P$ is a point on arc $BC$ that does not contain $ A$ such that $AP$ is the symmedian of triangle $ABC$. $E ,F$ are symmetric of $P$ wrt $CA, AB$ respectively . $K$ is symmetric of $A$ wrt $EF$. $L$ is the projection of $K$ on the line passing through $A$ and parallel to $BC$. Prove that $PA=PL$.

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 $\overline{AB}$ 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 $k$ 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 an acute triangle with circumcenter $O$ such that $AB=4$, $AC=5$, and $BC=6$. Let $D$ be the foot of the altitude from $A$ to $BC$, and $E$ be the intersection of $AO$ with $BC$. Suppose that $X$ is on $BC$ between $D$ and $E$ such that there is a point $Y$ on $AD$ satisfying $XY\parallel AO$ and $YO\perp AX$. Determine the length of $BX$.

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.

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 an acute, scalene triangle, and let $ M$, $ N$, and $ P$ be the midpoints of $ \overline{BC}$, $ \overline{CA}$, and $ \overline{AB}$, respectively. Let the perpendicular bisectors of $ \overline{AB}$ and $ \overline{AC}$ intersect ray $ AM$ in points $ D$ and $ E$ respectively, and let lines $ BD$ and $ CE$ intersect in point $ F$, inside of triangle $ ABC$. Prove that points $ A$, $ N$, $ F$, and $ P$ all lie on one circle.

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.

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 triangle with circumcircle $\Gamma$ and circumcentre $O$. Denote by $M$ the midpoint of $BC$. The point $D$ is the reflection of $A$ over $BC$, and the point $E$ is the intersection of $\Gamma$ and the ray $MD$. Let $S$ be the circumcentre of the triangle $ADE$. Prove that the points $A$, $E$, $M$, $O$, and $S$ lie on the same circle.

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 $ABC$ 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 $C$.

Let $BB_1$ be the symmedian of a nonisosceles acute-angled triangle $ABC$. Ray $BB_1$ meets the circumcircle of $ABC$ for the second time at point $L$. Let $AH_A, BH_B, CH_C$ be the altitudes of triangle $ABC$. Ray $BH_B$ meets the circumcircle of $ABC$ for the second time at point $T$. Prove that $H_A, H_C, T, L$ are concyclic.

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 $ABC$.

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.

For $\triangle ABC$ and its circumcircle $\omega$, draw the tangents at $B,C$ to $\omega$ meeting at $D$. Let the line $AD$ meet the circle with center $D$ and radius $DB$ at $E$ inside $\triangle ABC$. Let $F$ be the point on the extension of $EB$ and $G$ be the point on the segment $EC$ such that $\angle AFB=\angle AGE=\angle A$. Prove that the tangent at $A$ to the circumcircle of $\triangle AFG$ is parallel to $BC$. [i]Proposed by 61plus[/i]

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