$BB_{1}$ is a bisector of an acute triangle $ABC$. A perpendicular from $B_{1}$ to $BC$ meets a smaller arc $BC$ of a circumcircle of $ABC$ in a point $K$. A perpendicular from $B$ to $AK$ meets $AC$ in a point $L$. $BB_{1}$ meets arc $AC$ in $T$. Prove that $K$, $L$, $T$ are collinear. [i]V. Astakhov[/i]

Let $ABCD$ be a cyclic quadrilateral such that $AB = AD + BC$ and $CD < AB$. The diagonals $AC$ and $BD$ intersect at $P$, while the lines $AD$ and $BC$ intersect at $Q$. The angle bisector of $\angle APB$ meets $AB$ at $T$. Show that the circumcenter of the triangle $CTD$ lies on the circumcircle of the triangle $CQD$. [i]Proposed by Nikola Velov[/i]

In triangle $ ABC$, sides $ a,b$ and $ c$ are opposite angles $ A,B$ and $ C$ respectively. $ AD$ bisects angle $ A$ and meets $ BC$ at $ D$. Then if $ x \equal{} \overline{CD}$ and $ y \equal{} \overline{BD}$ the correct proportion is: $ \textbf{(A)}\ \frac {x}{a} \equal{} \frac {a}{b \plus{} c} \qquad\textbf{(B)}\ \frac {x}{b} \equal{} \frac {a}{a \plus{} c} \qquad\textbf{(C)}\ \frac {y}{c} \equal{} \frac {c}{b \plus{} c} \\ \textbf{(D)}\ \frac {y}{c} \equal{} \frac {a}{b \plus{} c} \qquad\textbf{(E)}\ \frac {x}{y} \equal{} \frac {c}{b}$

Let $\triangle ABC$ be an acute angled triangle and let $k$ be its circumscribed circle. A point $O$ is given in the interior of the triangle, such that $CE=CF$, where $E$ and $F$ are on $k$ and $E$ lies on $AO$ while $F$ lies on $BO$. Prove that $O$ is on the angle bisector of $\angle ACB$ if and only if $AC=BC$.

Let ABC be a triangle with $AB = 8$, $BC = 9$, and $CA = 10$. The line tangent to the circumcircle of $ABC$ at $A$ intersects the line $BC$ at $T$, and the circle centered at $T$ passing through $A$ intersects the line $AC$ for a second time at $S$. If the angle bisector of $\angle SBA$ intersects $SA$ at $P$, compute the length of segment $SP$. [i]2016 CCA Math Bonanza Team #9[/i]

Let ABCD be a parallelogram, M the midpoint of AB and N the intersection of CD and the angle bisector of ABC. Prove that CM and BN are perpendicular iff AN is the angle bisector of DAB.

Let $ABC$ be a triangle with $\angle CAB=60^{\circ}$ and with incenter $I$. Let points $D,E$ be on sides $AC,AB$, respectively, such that $BD$ and $CE$ are angle bisectors of angles $\angle ABC$ and $\angle BCA$, respectively. Show that $ID=IE$.

Two circles $ \Omega_{1}$ and $ \Omega_{2}$ are externally tangent to each other at a point $ I$, and both of these circles are tangent to a third circle $ \Omega$ which encloses the two circles $ \Omega_{1}$ and $ \Omega_{2}$. The common tangent to the two circles $ \Omega_{1}$ and $ \Omega_{2}$ at the point $ I$ meets the circle $ \Omega$ at a point $ A$. One common tangent to the circles $ \Omega_{1}$ and $ \Omega_{2}$ which doesn't pass through $ I$ meets the circle $ \Omega$ at the points $ B$ and $ C$ such that the points $ A$ and $ I$ lie on the same side of the line $ BC$. Prove that the point $ I$ is the incenter of triangle $ ABC$. [i]Alternative formulation.[/i] Two circles touch externally at a point $ I$. The two circles lie inside a large circle and both touch it. The chord $ BC$ of the large circle touches both smaller circles (not at $ I$). The common tangent to the two smaller circles at the point $ I$ meets the large circle at a point $ A$, where the points $ A$ and $ I$ are on the same side of the chord $ BC$. Show that the point $ I$ is the incenter of triangle $ ABC$.

Let $ABCD$ be a rectangle of centre $O$, such that $\angle DAC=60^{\circ}$. The angle bisector of $\angle DAC$ meets $DC$ at $S$. Lines $OS$ and $AD$ meet at $L$, and lines $BL$ and $AC$ meet at $M$. Prove that lines $SM$ and $CL$ are parallel.

As shown in the figure below, the in-circle of $ABC$ is tangent to sides $AB$ and $AC$ at $D$ and $E$ respectively, and $O$ is the circumcenter of $BCI$. Prove that $\angle ODB = \angle OEC$. [asy]import graph; size(5.55cm); pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-5.76,xmax=4.8,ymin=-3.69,ymax=3.71; pen zzttqq=rgb(0.6,0.2,0), wwwwqq=rgb(0.4,0.4,0), qqwuqq=rgb(0,0.39,0); pair A=(-2,2.5), B=(-3,-1.5), C=(2,-1.5), I=(-1.27,-0.15), D=(-2.58,0.18), O=(-0.5,-2.92); D(A--B--C--cycle,zzttqq); D(arc(D,0.25,-104.04,-56.12)--(-2.58,0.18)--cycle,qqwuqq); D(arc((-0.31,0.81),0.25,-92.92,-45)--(-0.31,0.81)--cycle,qqwuqq); D(A--B,zzttqq); D(B--C,zzttqq); D(C--A,zzttqq); D(CR(I,1.35),linewidth(1.2)+dotted+wwwwqq); D(CR(O,2.87),linetype("2 2")+blue); D(D--O); D((-0.31,0.81)--O); D(A); D(B); D(C); D(I); D(D); D((-0.31,0.81)); D(O); MP( "A", A, dir(110)); MP("B", B, dir(140)); D("C", C, dir(20)); D("D", D, dir(150)); D("E", (-0.31, 0.81), dir(60)); D("O", O, dir(290)); D("I", I, dir(100)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]

The diagonals $ AC$ and $ BD$ of a convex quadrilateral $ ABCD$ intersect at point $ M$. The bisector of $ \angle ACD$ meets the ray $ BA$ at $ K$. Given that $ MA \cdot MC \plus{}MA \cdot CD \equal{} MB \cdot MD$, prove that $ \angle BKC \equal{} \angle CDB$.

A point $P$ lies on the internal angle bisector of $\angle BAC$ of a triangle $\triangle ABC$. Point $D$ is the midpoint of $BC$ and $PD$ meets the external angle bisector of $\angle BAC$ at point $E$. If $F$ is the point such that $PAEF$ is a rectangle then prove that $PF$ bisects $\angle BFC$ internally or externally.

Given is the non-equilateral triangle $A_1A_2A_3$. $B_{ij}$ is the symmetric of $A_i$ wrt the inner bisector of $\angle A_j$. Prove that lines $B_{12}B_{21}$, $B_{13}B_{31}$ and $B_{23}B_{32}$ are parallel.

For a $\triangle ABC$ , let $A_1$ be the symmetric point of the intersection of angle bisector of $\angle BAC$ and $BC$ , where center of the symmetry is the midpoint of side $BC$, In the same way we define $B_1 $ ( on $AC$ ) and $C_1$ (on $AB$). Intersection of circumcircle of $\triangle A_1B_1C_1$ and line $AB$ is the set $\{Z,C_1 \}$, with $BC$ is the set $\{X,A_1\}$ and with $CA$ is the set $\{Y,B_1\}$. If the perpendicular lines from $X,Y,Z$ on $BC,CA$ and $ AB$ , respectively are concurrent , prove that $\triangle ABC$ is isosceles.

In a circle, let $AB$ and $BC$ be chords , with $AB =\sqrt3, BC =3\sqrt3, \angle ABC =60^o$. Find the length of the circle chord that divides angle $ \angle ABC$ in half.