In isosceles triangle $ABC(AC=BC)$ the point $M$ is in the segment $AB$ such that $AM=2MB,$ $F$ is the midpoint of $BC$ and $H$ is the orthogonal projection of $M$ in $AF.$ Prove that $\angle BHF=\angle ABC.$

$(BEL 3)$ Construct the circle that is tangent to three given circles.

Let $ABC$ an acute triangle and $H$ its orthocenter. Let $E$ and $F$ be the intersection of lines $BH$ and $CH$ with $AC$ and $AB$ respectively, and let $D$ be the intersection of lines $EF$ and $BC$. Let $\Gamma_1$ be the circumcircle of $AEF$, and $\Gamma_2$ the circumcircle of $BHC$. The line $AD$ intersects $\Gamma_1$ at point $I \neq A$. Let $J$ be the feet of the internal bisector of $\angle{BHC}$ and $M$ the midpoint of the arc $\stackrel{\frown}{BC}$ from $\Gamma_2$ that contains the point $H$. The line $MJ$ intersects $\Gamma_2$ at point $N \neq M$. Show that the triangles $EIF$ and $CNB$ are similar.

Consider a tetrahedron $SABC$. The incircle of the triangle $ABC$ has the center $I$ and touches its sides $BC$, $CA$, $AB$ at the points $E$, $F$, $D$, respectively. Let $A^{\prime}$, $B^{\prime}$, $C^{\prime}$ be the points on the segments $SA$, $SB$, $SC$ such that $AA^{\prime}=AD$, $BB^{\prime}=BE$, $CC^{\prime}=CF$, and let $S^{\prime}$ be the point diametrically opposite to the point $S$ on the circumsphere of the tetrahedron $SABC$. Assume that the line $SI$ is an altitude of the tetrahedron $SABC$. Show that $S^{\prime}A^{\prime}=S^{\prime}B^{\prime}=S^{\prime}C^{\prime}$.

In isosceles triangle $ABC(AC=BC)$ the point $M$ is in the segment $AB$ such that $AM=2MB,$ $F$ is the midpoint of $BC$ and $H$ is the orthogonal projection of $M$ in $AF.$ Prove that $\angle BHF=\angle ABC.$

In triangle $ABC$, $P$ is a point on altitude $AD$. $Q,R$ are the feet of the perpendiculars from $P$ to $AB,AC$, and $QP,RP$ meet $BC$ at $S$ and $T$ respectively. the circumcircles of $BQS$ and $CRT$ meet $QR$ at $X,Y$. a) Prove $SX,TY, AD$ are concurrent at a point $Z$. b) Prove $Z$ is on $QR$ iff $Z=H$, where $H$ is the orthocenter of $ABC$. [i]Ray Li.[/i]

Let $ AB$ be a diameter of a circle $ S$, and let $ L$ be the tangent at $ A$. Furthermore, let $ c$ be a fixed, positive real, and consider all pairs of points $ X$ and $ Y$ lying on $ L$, on opposite sides of $ A$, such that $ |AX|\cdot |AY| \equal{} c$. The lines $ BX$ and $ BY$ intersect $ S$ at points $ P$ and $ Q$, respectively. Show that all the lines $ PQ$ pass through a common point.

$ABCD$ is a convex quadrilateral; half-lines $DA$ and $CB$ meet at point $Q$; half-lines $BA$ and $CD$ meet at point $P$. It is known that $\angle AQB=\angle APD$. The bisector of angle $\angle AQB$ meets the sides $AB$ and $CD$ of the quadrilateral at points $X$ and $Y$, respectively; the bisector of angle $\angle APD$ meets the sides $AD$ and $BC$ at points $Z$ and $T$, respectively. The circumcircles of triangles $ZQT$ and $XPY$ meet at point $K$ inside the quadrilateral. Prove that $K$ lies on the diagonal $AC$. [i]Proposed by S. Berlov[/i]

Given a circle $(O)$ and two fixed points $B,C$ on $(O),$ and an arbitrary point $A$ on $(O)$ such that the triangle $ABC$ is acute. $M$ lies on ray $AB,$ $N$ lies on ray $AC$ such that $MA=MC$ and $NA=NB.$ Let $P$ be the intersection of $(AMN)$ and $(ABC),$ $P\ne A.$ $MN$ intersects $BC$ at $Q.$ a) Prove that $A,P,Q$ are collinear. b) $D$ is the midpoint of $BC.$ Let $K$ be the intersection of $(M,MA)$ and $(N,NA),$ $K\ne A.$ $d$ is the line passing through $A$ and perpendicular to $AK.$ $E$ is the intersection of $d$ and $BC.$ $(ADE)$ intersects $(O)$ at $F,$ $F\ne A.$ Prove that $AF$ passes through a fixed point.

An isosceles triangle with $AB=AC$ has an inscribed circle $O$, which touches its sides $BC,CA,AB$ at $K,L,M$ respectively. The lines $OL$ and $KM$ intersect at $N$; the lines $BN$ and $CA$ intersect at $Q$. Let $P$ be the foot of the perpendicular from $A$ on $BQ$. Suppose that $BP=AP+2\cdot PQ$. Then, what values can the ratio $\frac{AB}{BC}$ assume?

Let $ABC$ be an acute scalene triangle inscribed in circle $\Omega$. Circle $\omega$, centered at $O$, passes through $B$ and $C$ and intersects sides $AB$ and $AC$ at $E$ and $D$, respectively. Point $P$ lies on major arc $BAC$ of $\Omega$. Prove that lines $BD, CE, OP$ are concurrent if and only if triangles $PBD$ and $PCE$ have the same incenter.

Let $ABCD$ be a convex quadrilateral with $AB$ is not parallel to $CD$. Circle $\omega_1$ with center $O_1$ passes through $A$ and $B$, and touches segment $CD$ at $P$. Circle $\omega_2$ with center $O_2$ passes through $C$ and $D$, and touches segment $AB$ at $Q$. Let $E$ and $F$ be the intersection of circles $\omega_1$ and $\omega_2$. Prove that $EF$ bisects segment $PQ$ if and only if $BC$ is parallel to $AD$.

Let $\gamma$ be the incircle in the triangle $A_0A_1A_2$. For all $i\in\{0,1,2\}$ we make the following constructions (all indices are considered modulo 3): $\gamma_i$ is the circle tangent to $\gamma$ which passes through the points $A_{i+1}$ and $A_{i+2}$; $T_i$ is the point of tangency between $\gamma_i$ and $\gamma$; finally, the common tangent in $T_i$ of $\gamma_i$ and $\gamma$ intersects the line $A_{i+1}A_{i+2}$ in the point $P_i$. Prove that a) the points $P_0$, $P_1$ and $P_2$ are collinear; b) the lines $A_0T_0$, $A_1T_1$ and $A_2T_2$ are concurrent.

Let $ABCD$ be a cyclic quadrilateral with center $O$. Suppose the circumcircles of triangles $AOB$ and $COD$ meet again at $G$, while the circumcircles of triangles $AOD$ and $BOC$ meet again at $H$. Let $\omega_1$ denote the circle passing through $G$ as well as the feet of the perpendiculars from $G$ to $AB$ and $CD$. Define $\omega_2$ analogously as the circle passing through $H$ and the feet of the perpendiculars from $H$ to $BC$ and $DA$. Show that the midpoint of $GH$ lies on the radical axis of $\omega_1$ and $\omega_2$. [i]Proposed by Yang Liu[/i]

Let $ABC$ be a triangle, and let $D$, $E$, $F$ be the points of contact of its incircle $\omega$ with its sides $BC$, $CA$, $AB$, respectively. Let $K$ be the point of intersection of the line $AD$ with the incircle $\omega$ different from $D$, and let $M$ be the point of intersection of the line $EF$ with the line perpendicular to $AD$ passing through $K$. Prove that $AM$ is parallel to $BC$.

In $\triangle ABC$, $H$ is the orthocenter, and $AD,BE$ are arbitrary cevians. Let $\omega_1, \omega_2$ denote the circles with diameters $AD$ and $BE$, respectively. $HD,HE$ meet $\omega_1,\omega_2$ again at $F,G$. $DE$ meets $\omega_1,\omega_2$ again at $P_1,P_2$ respectively. $FG$ meets $\omega_1,\omega_2$ again $Q_1,Q_2$ respectively. $P_1H,Q_1H$ meet $\omega_1$ at $R_1,S_1$ respectively. $P_2H,Q_2H$ meet $\omega_2$ at $R_2,S_2$ respectively. Let $P_1Q_1\cap P_2Q_2 = X$, and $R_1S_1\cap R_2S_2=Y$. Prove that $X,Y,H$ are collinear. [i]Ray Li.[/i]

Let $ABC$ be a triangle with $ \widehat{BAC}\neq 90^\circ $. Let $M$ be the midpoint of $BC$. We choose a variable point $D$ on $AM$. Let $(O_1)$ and $(O_2)$ be two circle pass through $ D$ and tangent to $BC$ at $B$ and $C$. The line $BA$ and $CA$ intersect $(O_1),(O_2)$ at $ P,Q$ respectively. [b]a)[/b] Prove that tangent line at $P$ on $(O_1)$ and $Q$ on $(O_2)$ must intersect at $S$. [b]b)[/b] Prove that $S$ lies on a fix line.

Let $ABC$ be a triangle and $X,Y,Z$ be the tangency points of its inscribed circle with the sides $BC, CA, AB$, respectively. Suppose that $C_1, C_2, C_3$ are circle with chords $YZ, ZX, XY$, respectively, such that $C_1$ and $C_2$ intersect on the line $CZ$ and that $C_1$ and $C_3$ intersect on the line $BY$. Suppose that $C_1$ intersects the chords $XY$ and $ZX$ at $J$ and $M$, respectively; that $C_2$ intersects the chords $YZ$ and $XY$ at $L$ and $I$, respectively; and that $C_3$ intersects the chords $YZ$ and $ZX$ at $K$ and $N$, respectively. Show that $I, J, K, L, M, N$ lie on the same circle.

Two circles in the plane do not intersect and do not lie inside each other. We choose diameters $A_1B_1$ and $A_2B_2$ of these circles such that the segments $A_1A_2$ and $B_1B_2'$ intersect. Let $A$ and $B$ be the midpoints of the segments $A_1A_2$ and $B_1B_2$, and $C$ be the intersection point of these segments. Prove that the orthocenter of the triangle $ABC$ belongs to a fixed line that does not depend on the choice of diameters.

Diagonals of a cyclic quadrilateral $ABCD$ intersect at point $K.$ The midpoints of diagonals $AC$ and $BD$ are $M$ and $N,$ respectively. The circumscribed circles $ADM$ and $BCM$ intersect at points $M$ and $L.$ Prove that the points $K ,L ,M,$ and $ N$ lie on a circle. (all points are supposed to be different.)

The lines \(AB\) and \(CD\), containing the lateral sides of the trapezoid \(ABCD\), intersect at point \(Q\). Inside the trapezoid \(ABCD\), a point \(P\) is chosen such that \(\angle APB = \angle CPD\). Prove that the circumcircles of triangles \(BPD\) and \(APC\) intersect again on the line \(PQ\). [i]Proposed by Mykhailo Shtandenko[/i]

Let $P$ be a point inside the triangle $ABC$, such that the angles $\angle CBP$ and $\angle PAC$ are equal. Denote the intersection of the line $AP$ and the segment $BC$ by $D$, and the intersection of the line $BP$ with the segment $AC$ by $E$. The circumcircles of the triangles $ADC$ and $BEC$ meet at $C$ and $F$. Show that the line $CP$ bisects the angle $DFE$. Please remember to hide your solution. (by using the hide tags of course.. I don't literally mean that you should hide it :ninja: )

Octagon $A_1A_2A_3A_4A_5A_6A_7A_8$ is inscribed in a circle $\Omega$ with center $O$. It is known that $A_1A_2\|A_5A_6$, $A_3A_4\|A_7A_8$ and $A_2A_3\|A_5A_8$. The circle $\omega_{12}$ passes through $A_1$, $A_2$ and touches $A_1A_6$; circle $\omega_{34}$ passes through $A_3$, $A_4$ and touches $A_3A_8$; the circle $\omega_{56}$ passes through $A_5$, $A_6$ and touches $A_5A_2$; the circle $\omega_{78}$ passes through $A_7$, $A_8$ and touches $A_7A_4$. The common external tangent to $\omega_{12}$ and $\omega_{34}$ cross the line passing through ${A_1A_6}\cap{A_3A_8}$ and ${A_5A_2}\cap{A_7A_4}$ at the point $X$. Prove that one of the common tangents to $\omega_{56}$ and $\omega_{78}$ passes through $X$.

$ABC$ is an acute triangle. $P$(different from $B,C$) is a point on side $BC$. $H$ is an orthocenter, and $D$ is a foot of perpendicular from $H$ to $AP$. The circumcircle of the triangle $ABD$ and $ACD$ is $O _1$ and $O_2$, respectively. A line $l$ parallel to $BC$ passes $D$ and meet $O_1$ and $O_2$ again at $X$ and $Y$, respectively. $l$ meets $AB$ at $E$, and $AC$ at $F$. Two lines $XB$ and $YC$ intersect at $Z$. Prove that $ZE=ZF$ is a necessary and sufficient condition for $BP=CP$.

Point $A$ is outside of a given circle $\omega$. Let the tangents from $A$ to $\omega$ meet $\omega$ at $S, T$ points $X, Y$ are midpoints of $AT, AS$ let the tangent from $X$ to $\omega$ meet $\omega$ at $R\neq T$. points $P, Q$ are midpoints of $XT, XR$ let $XY\cap PQ=K, SX\cap TK=L$ prove that quadrilateral $KRLQ$ is cyclic.