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.

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]

A convex quadrilateral $ABCD$ satisfies $AB\cdot CD = BC\cdot DA$. Point $X$ lies inside $ABCD$ so that \[\angle{XAB} = \angle{XCD}\quad\,\,\text{and}\quad\,\,\angle{XBC} = \angle{XDA}.\] Prove that $\angle{BXA} + \angle{DXC} = 180^\circ$. [i]Proposed by Tomasz Ciesla, Poland[/i]

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 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 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 $A'B'$.

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 $ABCD$ be a parallelogram. The tangent to the circumcircle of triangle $BCD$ at $C$ intersects $AB$ at $P$ and intersects $AD$ at $Q$. The tangents to the circumcircle of triangle $APQ$ at $P$ and $Q$ meet at $R$. Show that points $A$, $C$, and $R$ are collinear.

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 $ABC$ 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 $k$.

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 $PA=PL$.

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

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]

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