In the acute-angled triangle $ ABC$, $ D$ is the intersection of the altitude passing through $ A$ with $ BC$ and $ I_a$ is the excenter of the triangle with respect to $ A$. $ K$ is a point on the extension of $ AB$ from $ B$, for which $ \angle AKI_a\equal{}90^\circ\plus{}\frac 34\angle C$. $ I_aK$ intersects the extension of $ AD$ at $ L$. Prove that $ DI_a$ bisects the angle $ \angle AI_aB$ iff $ AL\equal{}2R$. ($ R$ is the circumradius of $ ABC$)

let the incircle of a triangle ABC touch BC,AC,AB at A1,B1,C1 respectively. M and N are the midpoints of AB1 and AC1 respectively. MN meets A1C1 at T . draw two tangents TP and TQ through T to incircle. PQ meets MN at L and B1C1 meets PQ at K . assume I is the center of the incircle . prove IK is parallel to AL

Let $ABC$ be a triangle with $AB>AC$. Let $D$ be the foot of the internal angle bisector of $A$. Points $F$ and $E$ are on $AC,AB$ respectively such that $B,C,F,E$ are concyclic. Prove that the circumcentre of $DEF$ is the incentre of $ABC$ if and only if $BE+CF=BC$.

Let $O$ be the circumcenter of the triangle $ABC, A'$ be a point symmetric of $A$ wrt line $BC, X$ is an arbitrary point on the ray $AA'$ ($X \ne A$). Angle bisector of angle $BAC$ intersects the circumcircle of triangle $ABC$ at point $D$ ($D \ne A$). Let $M$ be the midpoint of the segment $DX$. A line passing through point $O$ parallel to $AD$, intersects $DX$ at point $N$. Prove that angles $BAM$ and $CAN$ angles are equal.

A tetrahedron is such that the center of the its circumscribed sphere is inside the tetrahedron. Show that at least one of its edges has a size larger than or equal to the size of the edge of a regular tetrahedron inscribed in this same sphere.

We are given a cyclic quadrilateral $ABCD$ with a point $E$ on the diagonal $AC$ such that $AD=AE$ and $CB=CE$. Let $M$ be the center of the circumcircle $k$ of the triangle $BDE$. The circle $k$ intersects the line $AC$ in the points $E$ and $F$. Prove that the lines $FM$, $AD$ and $BC$ meet at one point. [i](4th Middle European Mathematical Olympiad, Individual Competition, Problem 3)[/i]

Given triangle $ ABC$ and its circumcircle $ \Gamma$, let $ M$ and $ N$ be the midpoints of arcs $ BC$ (that does not contain $ A$) and $ CA$ (that does not contain $ B$), repsectively. Let $ X$ be a variable point on arc $ AB$ that does not contain $ C$. Let $ O_1$ and $ O_2$ be the incenter of triangle $ XAC$ and $ XBC$, respectively. Let the circumcircle of triangle $ XO_1O_2$ meets $ \Gamma$ at $ Q$. (a) Prove that $ QNO_1$ and $ QMO_2$ are similar. (b) Find the locus of $ Q$ as $ X$ varies.

Let $ P$ be a point in triangle $ ABC$, and define $ \alpha,\beta,\gamma$ as follows: \[ \alpha\equal{}\angle BPC\minus{}\angle BAC, \quad \beta\equal{}\angle CPA\minus{}\angle \angle CBA, \quad \gamma\equal{}\angle APB\minus{}\angle ACB.\] Prove that \[ PA\dfrac{\sin \angle BAC}{\sin \alpha}\equal{}PB\dfrac{\sin \angle CBA}{\sin \beta}\equal{}PC\dfrac{\sin \angle ACB}{\sin \gamma}.\]

Let $ABC$ be a triangle and $D$ a point on the side $BC$. The tangent line to the circumcircle of the triangle $ABD$ at the point $D$ intersects the side $AC$ at $E$. The tangent line to the circumcircle of the triangle $ACD$ at the the point $D$ intersects the side $AB$ at $F$. Prove that the point $A$ and the circumcenters of the triangles $ABC$ and $DEF$ are collinear. (Malik Talbi)

Consider an acute scalene triangle $\triangle{ABC}$. The interior bisector of $A$ intersects $BC$ at $E$ and the minor arc of $\overarc {BC}$ in circumcircle of $\triangle{ABC}$ at $M$. Suppose that $D$ is a point on the minor arc of $\overarc {BC}$ such that $ED=EM$. $P$ is a point on the line segment of $AD$ such that $\angle ABP=\angle ACP \not= 0$. $O$ is the circumcenter of $\triangle{ABC}$. Prove that $OP \perp AM$.

Let $P$ be a point in the interior of the equilateral triangle $\triangle ABC$ such that $\sphericalangle{APC}=120^\circ$. Let $M$ be the intersection of $CP$ with $AB$, and $N$ the intersection of $AP$ and $BC$. Find the locus of the circumcentre of the triangle $MBN$ as $P$ varies.

Let triangle $ABC$ have $\angle BAC = 45^{\circ}$ and circumcircle $\Gamma$ and let $M$ be the intersection of the angle bisector of $\angle BAC$ with $\Gamma$. Let $\Omega$ be the circle tangent to segments $\overline{AB}$ and $\overline{AC}$ and internally tangent to $\Gamma$ at point $T$. Given that $\angle TMA = 45^{\circ}$ and that $TM = \sqrt{100 - 50 \sqrt{2}}$, the length of $BC$ can be written as $a \sqrt{b}$, where $b$ is not divisible by the square of any prime. Find $a + b$.

Let $H$ be the orthocentre of an acute triangle $ABC$. The line through $A$ perpendicular to $AC$ and the line through $B$ perpendicular to $BC$ intersect in $D$. The circle with centre $C$ through $H$ intersects the circumcircle of triangle $ABC$ in the points $E$ and $F$. Prove that $|DE| = |DF| = |AB|$.

Let $ABC$ an acute triangle and $\Gamma$ its circumcircle. The bisector of $BAC$ intersects $\Gamma$ at $M\neq A$. A line $r$ parallel to $BC$ intersects $AC$ at $X$ and $AB$ at $Y$. Also, $MX$ and $MY$ intersect $\Gamma$ again at $S$ and $T$, respectively. If $XY$ and $ST$ intersect at $P$, prove that $PA$ is tangent to $\Gamma$.

Let $ABCD$ be a circumscribed quadrilateral, assume $ABCD$ is not a kite. Denote the circumcenters of triangle $ABC,BCD,CDA,DAB$ by $O_D,O_A,O_B,O_C$ respectively. a. Prove that $O_AO_BO_CO_D$ is circumscribed. b. Let the angle bisector of $\angle BAD$ intersect the angle bisector of $\angle O_BO_AO_D$ in $X$. Similarly define the points $Y,Z,W$. Denote the incenters of $ABCD, O_AO_BO_CO_D$ by $I,J$ respectively. Express the angles $\angle ZYJ,\angle XYI$ in terms of angles of quadrilateral $ABCD$.

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]

In the triangle $ABC$, $\angle A$ is biggest. On the circumcircle of $\triangle ABC$, let $D$ be the midpoint of $\widehat{ABC}$ and $E$ be the midpoint of $\widehat{ACB}$. The circle $c_1$ passes through $A,B$ and is tangent to $AC$ at $A$, the circle $c_2$ passes through $A,E$ and is tangent $AD$ at $A$. $c_1$ and $c_2$ intersect at $A$ and $P$. Prove that $AP$ bisects $\angle BAC$. [hide="Diagram"][asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(14.4cm); 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 = -5.23, xmax = 9.18, ymin = -2.97, ymax = 4.82; /* image dimensions */ /* draw figures */ draw(circle((-1.32,1.36), 2.98)); draw(circle((3.56,1.53), 3.18)); draw((0.92,3.31)--(-2.72,-1.27)); draw(circle((0.08,0.25), 3.18)); draw((-2.72,-1.27)--(3.13,-0.65)); draw((3.13,-0.65)--(0.92,3.31)); draw((0.92,3.31)--(2.71,-1.54)); draw((-2.41,-1.74)--(0.92,3.31)); draw((0.92,3.31)--(1.05,-0.43)); /* dots and labels */ dot((-1.32,1.36),dotstyle); dot((0.92,3.31),dotstyle); label("$A$", (0.81,3.72), NE * labelscalefactor); label("$c_1$", (-2.81,3.53), NE * labelscalefactor); dot((3.56,1.53),dotstyle); label("$c_2$", (3.43,3.98), NE * labelscalefactor); dot((1.05,-0.43),dotstyle); label("$P$", (0.5,-0.43), NE * labelscalefactor); dot((-2.72,-1.27),dotstyle); label("$B$", (-3.02,-1.57), NE * labelscalefactor); dot((2.71,-1.54),dotstyle); label("$E$", (2.71,-1.86), NE * labelscalefactor); dot((3.13,-0.65),dotstyle); label("$C$", (3.39,-0.9), NE * labelscalefactor); dot((-2.41,-1.74),dotstyle); label("$D$", (-2.78,-2.07), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/hide]

Let $ABC$ be an acute triangle with $AB = AC$. Let $D$ be the point on $BC$ such that $AD$ is perpendicular to $BC$. Let $O,H,G$ be the circumcenter, orthocenter and centroid of triangle $ABC$ respectively. Suppose that $2 \cdot OD = 23 \cdot HD$. Prove that $G$ lies on the incircle of triangle $ABC$.

Let $BH$ be an altitude of a right-angled triangle $ABC$ ($\angle B = 90^o$). The incircle of triangle $ABH$ touches $AB,AH$ in points $H_1, B_1$, the incircle of triangle $CBH$ touches $CB,CH$ in points $H_2, B_2$, point $O$ is the circumcenter of triangle $H_1BH_2$. Prove that $OB_1 = OB_2$.

In the plane points $A,B,C$ are marked in blue and points $P,Q$ are marked in red (no 3 marked points lie on a line, and no 4 marked points lie on a circle). A circle is called [b]separating[/b] if all points of one color are inside it, and all points of the other color are outside of it. Denote by $O$ the circumcenter of $ABC$ and by $R$ the circumradius of $ABC$. Prove that [b]exactly one[/b] of the following holds: [list] [*] There exists a separating circle; [*] There exists a point $X$ on the segment $PQ$ which also lies inside the triangle $ABC$, for which $PX\cdot XQ = R^2-OX^2$.

Let $ \Omega$ be the circumcircle of triangle $ ABC$. Let $ D$ be the point at which the incircle of $ ABC$ touches its side $ BC$. Let $ M$ be the point on $ \Omega$ such that the line $ AM$ is parallel to $ BC$. Also, let $ P$ be the point at which the circle tangent to the segments $ AB$ and $ AC$ and to the circle $ \Omega$ touches $ \Omega$. Prove that the points $ P$, $ D$, $ M$ are collinear.

Let $ABC$ be an acute triangle, $H$ its orthocentre, $D$ a point on the side $[BC]$, and $P$ a point such that $ADPH$ is a parallelogram. Show that $\angle BPC > \angle BAC$.

Given a rhombus $ABCD$, find the locus of the points $P$ lying inside the rhombus and satisfying $\angle APD+\angle BPC=180^{\circ}$.

The quadrilateral $ABCD$ is an isosceles trapezoid, where $AB\parallel CD$. The trapezoid is inscribed in a circle with radius $R$ and center on side $AB$. Point $E$ lies on the circumscribed circle and is such that $\angle DAE = 90^o$. Given that $\frac{AE}{AB}=\frac34$, calculate the length of the sides of the isosceles trapezoid.

Consider an acute triangle $ ABC. $ The points $ D,E,F $ are the feet of the altitudes of $ ABC $ from $ A,B,C, $ respectively. $ M,N,P $ are the middlepoints of $ BC,CA,AB, $ respectively. The circumcircles of $ BDP,CDN $ cross at $ X\neq D, $ the circumcircles of $ CEM,AEP $ cross at $ Y\neq E, $ and the circumcircles of $ AFN,BFM $ cross at $ Z\neq F. $ Prove that $ AX,BY,CZ $ are concurrent.