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]

Let $ABC$ be a triangle with incenter $I$, and let $D$ be a point on side $BC$. Points $X$ and $Y$ are chosen on lines $BI$ and $CI$ respectively such that $DXIY$ is a parallelogram. Points $E$ and $F$ are chosen on side $BC$ such that $AX$ and $AY$ are the angle bisectors of angles $\angle BAE$ and $\angle CAF$ respectively. Let $\omega$ be the circle tangent to segment $EF$, the extension of $AE$ past $E$, and the extension of $AF$ past $F$. Prove that $\omega$ is tangent to the circumcircle of triangle $ABC$.

Let $ABC$ be a triangle with circumcircle $\omega$. Let the bisector of $\angle ABC$ meet segment $AC$ at $D$ and circle $\omega$ at $M\ne B$. The circumcircle of $\triangle BDC$ meets line $AB$ at $E\ne B$, and $CE$ meets $\omega$ at $P\ne C$. The bisector of $\angle PMC$ meets segment $AC$ at $Q\ne C$. Given that $PQ = MC$, determine the degree measure of $\angle ABC$. [i]Ray Li.[/i]

Let $ ABC$ be a triangle, and let the angle bisectors of its angles $ CAB$ and $ ABC$ meet the sides $ BC$ and $ CA$ at the points $ D$ and $ F$, respectively. The lines $ AD$ and $ BF$ meet the line through the point $ C$ parallel to $ AB$ at the points $ E$ and $ G$ respectively, and we have $ FG \equal{} DE$. Prove that $ CA \equal{} CB$. [i]Original formulation:[/i] Let $ ABC$ be a triangle and $ L$ the line through $ C$ parallel to the side $ AB.$ Let the internal bisector of the angle at $ A$ meet the side $ BC$ at $ D$ and the line $ L$ at $ E$ and let the internal bisector of the angle at $ B$ meet the side $ AC$ at $ F$ and the line $ L$ at $ G.$ If $ GF \equal{} DE,$ prove that $ AC \equal{} BC.$

Let $ABC$ be an acute-angled triangle in which $|AB| < |AC|$ and let circle $k$ with center $O$ be its circumscribed circle. Let $P$ and $Q$ be points on sides $\overline{BC}$ and $\overline{AB}$ respectively such that $AQPO$ is a parallelogram. Let $K$ and $L$ be the intersections of the perpendicular bisector of $\overline{OP}$ with circle $k$, where $K$ is on the shorter arc $AB$. Let $M$ be the second intersection of the line $KQ$ with the circle $k$. Prove that the point $A$ belongs to the bisector of the angle $\angle QLM$.

A quadrilateral $ABCD$ is such that the sides $AB$ and $DC$ are parallel, and $|BC| =|AB| + |CD|$. Prove that the angle bisectors of the angles $\angle ABC$ and $\angle BCD$ intersect at right angles on the side $AD$.

$\vartriangle ABC$ has side lengths $AB = 4$ and $AC = 9$. Angle bisector $AD$ bisects angle $A$ and intersects $BC$ at $D$. Let $k$ be the ratio $\frac{BD}{AB}$ . Given that the length $AD$ is an integer, find the sum of all possible $k^2$ .

Prove the following two statements: (a) If a triangle is isosceles, then two of its bisectors are of equal length. (b) If two angle bisectors in a triangle are of equal length, then it is isosceles.

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.

The incenter of $\triangle A_1A_2A_3$ is $I$, and the center of the $A_i$-excircle is $J_i$ ($i=1,2,3$). Let $B_i$ be the intersection point of side $A_{i+1}A_{i+2}$ and the bisector of $\angle A_{i+1}IA_{i+2}$ ($A_{i+3}:=A_i$ $\forall i$). Prove that the three lines $B_iJ_i$ are concurrent.

Let $ABC$ be a triangle, $H$ the orthocenter and $M$ the midpoint of $AC$. Let $\ell$ be the parallel through $M$ to the bisector of $\angle AHC$. Prove that $\ell$ divides the triangle in two parts of equal perimeters. [i]Pedro Marrone, Panamá[/i]

In an acute triangle $ABC$ with $\overline{AB} < \overline{AC}$, angle bisector of $A$ and perpendicular bisector of $\overline{BC}$ intersect at $D$. Let $P$ be an interior point of triangle $ABC$. Line $CP$ meets the circumcircle of triangle $ABP$ again at $K$. Prove that $B, D, K$ are collinear if and only if $AD$ and $BC$ meet on the circumcircle of triangle $APC$.

Let angle bisectors of angles $\angle BAC$ and $\angle ABC$ of triangle $ABC$ intersect sides $BC$ and $AC$ in points $D$ and $E$, respectively. Let points $F$ and $G$ be foots of perpendiculars from point $C$ on lines $AD$ and $BE$, respectively. Prove that $FG \mid \mid AB$

Let $ABC$ be a triangle with $|AB|> |BC|$. Let $D$ be the midpoint of $AC$. Let $E$ be the intersection of the angular bisector of $\angle ABC$ and the line $AC$. Let $F$ be the point on $BE$ such that $CF$ is perpendicular to $BE$. Finally, let $G$ be the intersection of $CF$ and $BD$. Prove that $DF$ divides the line segment $EG$ into two equal parts.

The vertices $ A$ and $ B$ of an equilateral triangle $ ABC$ lie on a circle $k$ of radius $1$, and the vertex $ C$ is in the interior of the circle $ k$. A point $ D$, different from $ B$, lies on $ k$ so that $ AD\equal{}AB$. The line $ DC$ intersects $ k$ for the second time at point $ E$. Find the length of the line segment $ CE$.

Let $ABCD$ be a parallelogram. Let $M$ be a point on the side $AB$ and $N$ be a point on the side $BC$ such that the segments $AM$ and $CN$ have equal lengths and are non-zero. The lines $AN$ and $CM$ meet at $Q$. Prove that the line $DQ$ is the bisector of the angle $\measuredangle ADC$. [i]Alternative formulation.[/i] Let $ABCD$ be a parallelogram. Let $M$ and $N$ be points on the sides $AB$ and $BC$, respectively, such that $AM=CN\neq 0$. The lines $AN$ and $CM$ intersect at a point $Q$. Prove that the point $Q$ lies on the bisector of the angle $\measuredangle ADC$.

Side $\overline{AB}$ of $\triangle ABC$ has length $10$. The bisector of angle $A$ meets $\overline{BC}$ at $D$, and $CD = 3$. The set of all possible values of $AC$ is an open interval $(m,n)$. What is $m+n$? $ \textbf{(A) }16 \qquad \textbf{(B) }17 \qquad \textbf{(C) }18 \qquad \textbf{(D) }19 \qquad \textbf{(E) }20 \qquad $

Let $ABC$ be a triangle and $D$ point on side $BC$ such that $AD$ is angle bisector of angle $\angle BAC$. Let $E$ be the intersection of the side $AB$ with circle $\omega_1$ which has diameter $CD$ and let $F$ be the intersection of the side $AC$ with circle $\omega_2$ which has diameter $BD$. Suppose that there exist points $P\in\omega_1$ and $Q\in\omega_2$ such that $E, P, Q$ and $F$ are collinear and on this order. Prove that $AD, BQ$ and $CP$ are concurrent. [i]Proposed by Dorlir Ahmeti, Kosovo and Noah Walsh, U.S.A.[/i]

Let $ABC$ be an acute-angled triangle in which $\angle ABC$ is the largest angle. Let $O$ be its circumcentre. The perpendicular bisectors of $BC$ and $AB$ meet $AC$ at $X$ and $Y$ respectively. The internal angle bisectors of $\angle AXB$ and $\angle BYC$ meet $AB$ and $BC$ at $D$ and $E$ respectively. Prove that $BO$ is perpendicular to $AC$ if $DE$ is parallel to $AC$.

In $\triangle ABC$, let $AD$ be the angle bisector of $\angle BAC$, with $D$ on $BC$. The perpendicular from $B$ to $AD$ intersects the circumcircle of $\triangle ABD$ at $B$ and $E$. Prove that $E$, $A$ and the circumcenter $O$ of $\triangle ABC$ are collinear.

In a triangle $ABC, L$ and $K$ are the points of intersections of the angle bisectors of $\angle ABC$ and $\angle BAC$ with the segments $AC$ and $BC$, respectively. The segment $KL$ is angle bisector of $\angle AKC$, determine $\angle BAC$.

Let $ABC$ be a triangle with $\angle BAC = 60^{\circ}$. Let $AP$ bisect $\angle BAC$ and let $BQ$ bisect $\angle ABC$, with $P$ on $BC$ and $Q$ on $AC$. If $AB + BP = AQ + QB$, what are the angles of the triangle?

Let $\triangle ABC$ with $\angle BAC=120 ^\circ$. Let $D, E, F$ points on sides $BC, CA, AB$, respectively, such that $AD, BE, CF$ are angle bisectors on $\triangle ABC$. \\ Prove that $\triangle ABC$ is isosceles if and only if $\triangle DEF$ is right-angled isosceles.

Let $ABC$ be a triangle inscribed in the circle $(O)$. Let $I$ be the center of the circle inscribed in the triangle and $D$ the point of contact of the circle inscribed with the side $BC$. Let $M$ be the second intersection point of the bisector $AI$ with the circle $(O)$ and let $P$ be the point where the line $DM$ intersects the circle $(O)$ . Show that $PA \perp PI$.

In rectangle $ABCD$, $AB=20$ and $BC=10$. Let $E$ be a point on $\overline{CD}$ such that $\angle CBE=15^\circ$. What is $AE$? $ \textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20 $