The angle $B$ of a triangle $ABC$ is $120^o$. Let $M$ be a point on the side $AC$ and $K$ a point on the extension of the side $AB$, such that $BM$ is the internal bisector of the angle $\angle ABC$ and $CK$ is the external bisector corresponding to the angle $\angle ACB$ . The segment $MK$ intersects $BC$ at point $P$. Prove that $\angle APM = 30^o$.

Given a triangle $ABC$, let $D$ be a point different from $A$ on the external bisectrix $\ell$ of the angle $BAC$, and let $E$ be an interior point of the segment $AD$. Reflect $\ell$ in the internal bisectrices of the angles $BDC$ and $BEC$ to obtain two lines that meet at some point $F$. Show that the angles $ABD$ and $EBF$ are congruent.

A circle $\omega$ with center $I$ is inscribed into a segment of the disk, formed by an arc and a chord $AB$. Point $M$ is the midpoint of this arc $AB$, and point $N$ is the midpoint of the complementary arc. The tangents from $N$ touch $\omega$ in points $C$ and $D$. The opposite sidelines $AC$ and $BD$ of quadrilateral $ABCD$ meet in point $X$, and the diagonals of $ABCD$ meet in point $Y$. Prove that points $X, Y, I$ and $M$ are collinear.

We are given a circle $k$ and a point $A$ outside of $k$. Next we draw three lines through $A$: one secant intersecting the circle $k$ at points $B$ and $C$, and two tangents touching the circle$k$ at points $D$ and $E$. Let $F$ be the midpoint of $DE$. Show that the line $DE$ bisects the angle $\angle BFC$.

Let $\triangle ABC$ be a triangle with $|AC|=2|AB|$ and let $O$ be its circumcenter. Let $D$ be the intersection of the bisector of $\angle A$ with $BC$. Let $E$ be the orthogonal projection of $O$ to $AD$ and let $F\ne D$ be the point on $AD$ satisfying $|CD|=|CF|$. Prove that $\angle EBF=\angle ECF$.

Let $ABC$ be an acute triangle and $A_{1}, B_{1}, C_{1}$ be the touchpoints of the excircles with the segments $BC, CA, AB$ respectively. Let $O_{A}, O_{B}, O_{C}$ be the circumcenters of $\triangle AB_{1}C_{1}, \triangle BC_{1}A_{1}, \triangle CA_{1}B_{1}$ respectively. Prove that the lines through $O_{A}, O_{B}, O_{C}$ respectively parallel to the internal angle bisectors of $\angle A,\angle B, \angle C$ are concurrent.

The side lengths of $\triangle ABC$ are integers with no common factor greater than $1$. Given that $\angle B = 2 \angle C$ and $AB < 600$, compute the sum of all possible values of $AB$. [i]Proposed by Eugene Chen[/i]

In the diagram below, angle $ABC$ is a right angle. Point $D$ is on $\overline{BC}$, and $\overline{AD}$ bisects angle $CAB$. Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$, respectively, so that $AE=3$ and $AF=10.$ Given that $EB=9$ and $FC=27$, find the integer closest to the area of quadrilateral $DCFG.$ [asy] size(250); pair A=(0,12), E=(0,8), B=origin, C=(24*sqrt(2),0), D=(6*sqrt(2),0), F=A+10*dir(A--C), G=intersectionpoint(E--F, A--D); draw(A--B--C--A--D^^E--F); pair point=G+1*dir(250); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$G$", G, dir(point--G)); markscalefactor=0.1; draw(rightanglemark(A,B,C)); label("10", A--F, dir(90)*dir(A--F)); label("27", F--C, dir(90)*dir(F--C)); label("3", (0,10), W); label("9", (0,4), W);[/asy]

Let $A_{1}$, $B_{1}$, $C_{1}$ be the feet of the altitudes of an acute-angled triangle $ABC$ issuing from the vertices $A$, $B$, $C$, respectively. Let $K$ and $M$ be points on the segments $A_{1}C_{1}$ and $B_{1}C_{1}$, respectively, such that $\measuredangle KAM = \measuredangle A_{1}AC$. Prove that the line $AK$ is the angle bisector of the angle $C_{1}KM$.

Triangle $ABC$ has $AC = 3$, $BC = 5$, $AB = 7$. A circle is drawn internally tangent to the circumcircle of $ABC$ at $C$, and tangent to $AB$. Let $D$ be its point of tangency with $AB$. Find $BD - DA$. [asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(6cm); real labelscalefactor = 2.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.5, xmax = 7.01, ymin = -3, ymax = 8.02; /* image dimensions */ /* draw figures */ draw(circle((1.37,2.54), 5.17)); draw((-2.62,-0.76)--(-3.53,4.2)); draw((-3.53,4.2)--(5.6,-0.44)); draw((5.6,-0.44)--(-2.62,-0.76)); draw(circle((-0.9,0.48), 2.12)); /* dots and labels */ dot((-2.62,-0.76),dotstyle); label("$C$", (-2.46,-0.51), SW * labelscalefactor); dot((-3.53,4.2),dotstyle); label("$A$", (-3.36,4.46), NW * labelscalefactor); dot((5.6,-0.44),dotstyle); label("$B$", (5.77,-0.17), SE * labelscalefactor); dot((0.08,2.37),dotstyle); label("$D$", (0.24,2.61), SW * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); label("$7$",(-3.36,4.46)--(5.77,-0.17), NE * labelscalefactor); label("$3$",(-3.36,4.46)--(-2.46,-0.51),SW * labelscalefactor); label("$5$",(-2.46,-0.51)--(5.77,-0.17), SE * labelscalefactor); /* end of picture */ [/asy]

The angle bisector of vertex $A$ of $\triangle ABC$ cuts $[BC]$ at $D$. The circle passing through $A$ and touching to $BC$ at $D$ meets $[AB]$ and $[AC]$ at $P$ and $Q$, respectively. $AD$ and $PQ$ meet at $T$. If $|AB|=5, |BC|=6, |CA|=7$, then $\frac{|AT|}{|TD|}=?$ $ \textbf{(A)}\ \frac75 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ \frac72 \qquad \textbf{(E)}\ 4$

Point $D$ is on side $CB$ of triangle $ABC$. If \[ \angle{CAD} = \angle{DAB} = 60^\circ,\quad AC = 3\quad\mbox{ and }\quad AB = 6, \] then the length of $AD$ is $\text{(A)} \ 2 \qquad \text{(B)} \ 2.5 \qquad \text{(C)} \ 3 \qquad \text{(D)} \ 3.5 \qquad \text{(E)} \ 4$

Given a triangle $ABC$ where $AC > BC$, $D$ is located on the circumcircle of $ABC$ such that $D$ is the midpoint of the arc $AB$ that contains $C$. $E$ is a point on $AC$ such that $DE$ is perpendicular to $AC$. Prove that $AE = EC + CB$.

Let $ABC$ be an isosceles triangle such that $\angle BAC = 100^{\circ}$. Let $D$ be an intersection point of angle bisector of $\angle ABC$ and side $AC$, prove that $AD+DB=BC$

Three angle bisectors were drawn in a triangle, and it turned out that the angles between them are $50^o$, $60^o$ and $70^o$. Find the angles of the original triangle.

Denote $\overline{w_a}, \overline{w_b}, \overline{w_c}$ the external angle-bisectors in triangle $ABC$, prove that $$\sum \limits_{cyc} \frac{1}{w_a}\leq \sqrt{\frac{(s^2 - r^2 - 4Rr)(8R^2 - s^2 - r^2 - 2Rr)}{8s^2R^2r}}$$

Given a convex quadrilateral $ABCD$, such that $OA = \frac{OB \cdot OD}{OC + CD}$ where $O$ is the intersection of the two diagonals. The circumcircle of triangle $ABC$ intersects $BD$ at point $Q$. Prove that $CQ$ bisects $\angle ACD$

Let $AD$ and $BE$ be angle bisectors in triangle $ABC$. Let $x$, $y$ and $z$ be distances from point $M$, which lies on segment $DE$, from sides $BC$, $CA$ and $AB$, respectively. Prove that $z=x+y$

Let $ ABC$ be an isosceles right triangle and $M$ be the midpoint of its hypotenuse $AB$. Points $D$ and $E$ are taken on the legs $AC$ and $BC$ respectively such that $AD=2DC$ and $BE=2EC$. Lines $AE$ and $DM$ intersect at $F$. Show that $FC$ bisects the $\angle DFE$.

Two equal non-intersecting wooden disks, one gray and one black, are glued to a plane. A triangle with one gray side and one black side can be moved along the plane so that the disks remain outside the triangle, while the colored sides of the triangle are tangent to the disks of the same color (the tangency points are not the vertices). Prove that the line that contains the bisector of the angle between the gray and black sides always passes through some fixed point of the plane. (Egor Bakaev, Pavel Kozhevnikov, Vladimir Rastorguev) (Senior version[url=https://artofproblemsolving.com/community/c6h2102856p15209040] here[/url])

Let $H$, $I$, $O$ be the orthocenter, incenter and circumcenter of a triangle. Show that $2 \cdot IO \geq IH$. When does the equality hold ?

Let $ABC$ be a triangle with $\angle A = 80^o$ and $\angle C = 30^o$. Consider the point $M$ inside the triangle $ABC$ so that $\angle MAC= 60^o$ and $\angle MCA = 20^o$. If $N$ is the intersection of the lines $BM$ and $AC$ to show that a $MN$ is the bisector of the angle $\angle AMC$.

Consider the triangle $ABC$, where $\angle B= 30^o, \angle C = 15^o$, and $M$ is the midpoint of the side $[BC]$. Let point $N \in (BC)$ be such that $[NC] = [AB]$. Show that $[AN$ is the angle bisector of $MAC$

Let $E$ and $F$ be the intersections of opposite sides of a convex quadrilateral $ABCD$. The two diagonals meet at $P$. Let $O$ be the foot of the perpendicular from $P$ to $EF$. Show that $\angle BOC=\angle AOD$.

The incircle $\omega$ of a triangle $ABC$ is tangent to the sides $AB$ and $BC$ at $P$ and $Q$ respectively. The angle bisector at $A$ meets $PQ$ at point $S$. Prove $\angle ASC = 90^o$ .