Let $ ABC$ be a triangle and let $ P$ be a point on the angle bisector $ AD$, with $ D$ on $ BC$. Let $ E$, $ F$ and $ G$ be the intersections of $ AP$, $ BP$ and $ CP$ with the circumcircle of the triangle, respectively. Let $ H$ be the intersection of $ EF$ and $ AC$, and let $ I$ be the intersection of $ EG$ and $ AB$. Determine the geometric place of the intersection of $ BH$ and $ CI$ when $ P$ varies.

The incircle of triangle $ABC$ touches $BC$ at $D$ and $AC$ at $E$. Let $K$ be the point on $CB$ with $CK=BD$, and $L$ be the point on $CA$ with $AE=CL$. Lines $AK$ and $BL$ meet at $P$. If $Q$ is the midpoint of $BC$, $I$ the incenter, and $G$ the centroid of $\triangle ABC$, show that: $(a)$ $IQ$ and $AK$ are parallel, $(b)$ the triangles $AIG$ and $QPG$ have equal area.

Points $A$, $B$, $C$ and $D$ are chosen on a line in that order, with $AB$ and $CD$ greater than $BC$. Equilateral triangles $APB$, $BCQ$ and $CDR$ are constructed so that $P$, $Q$ and $R$ are on the same side with respect to $AD$. If $\angle PQR=120^\circ$, show that \[\frac{1}{AB}+\frac{1}{CD}=\frac{1}{BC}.\]

In triangle $ABC$, $AB=13$, $BC=14$ and $CA=15$. Segment $BC$ is split into $n+1$ congruent segments by $n$ points. Among these points are the feet of the altitude, median, and angle bisector from $A$. Find the smallest possible value of $n$. [i]Proposed by Evan Chen[/i]

Triangle $ ABC$ has $ AC \equal{} 450$ and $ BC \equal{} 300$. Points $ K$ and $ L$ are located on $ \overline{AC}$ and $ \overline{AB}$ respectively so that $ AK \equal{} CK$, and $ \overline{CL}$ is the angle bisector of angle $ C$. Let $ P$ be the point of intersection of $ \overline{BK}$ and $ \overline{CL}$, and let $ M$ be the point on line $ BK$ for which $ K$ is the midpoint of $ \overline{PM}$. If $ AM \equal{} 180$, find $ LP$.

Let $D$ and $E$ be foots of altitudes from $A$ and $B$ of triangle $ABC$, $F$ be intersection point of angle bisector from $C$ with side $AB$, and $O$, $I$ and $H$ be circumcenter, center of inscribed circle and orthocenter of triangle $ABC$, respectively. If $\frac{CF}{AD}+ \frac{CF}{BE}=2$, prove that $OI = IH$.

Let $ABC$ be a triangle such that it's circumcircle radius is equal to the radius of outer inscribed circle with respect to $A$. Suppose that the outer inscribed circle with respect to $A$ touches $BC,AC,AB$ at $M,N,L$. Prove that $O$ (Center of circumcircle) is the orthocenter of $MNL$.

In a triangle $ABC$, with $AB \ne BC$, $E$ is a point on the line $AC$ such that $BE$ is perpendicular to $AC$. A circle passing through $A$ and touching the line $BE$ at a point $P \ne B$ intersects the line $AB$ for the second time at $X$. Let $Q$ be a point on the line $PB$ different from $P$ such that $BQ = BP$. Let $Y$ be the point of intersection of the lines $CP$ and $AQ$. Prove that the points $C, X, Y, A$ are concyclic if and only if $CX$ is perpendicular to $AB$.

Let $ABCD$ be a rectangle and denote by $M$ and $N$ the midpoints of $AD$ and $BC$ respectively. The point $P$ is on $(CD$ such that $D\in (CP)$, and $PM$ intersects $AC$ in $Q$. Prove that $m(\angle{MNQ})=m(\angle{MNP})$.

Cirlce $\Omega$ is inscribed in triangle $ABC$ with $\angle BAC=40$. Point $D$ is inside the angle $BAC$ and is the intersection of exterior bisectors of angles $B$ and $C$ with the common side $BC$. Tangent form $D$ touches $\Omega$ in $E$. FInd $\angle BEC$.

It is given triangle $ABC$ and points $P$ and $Q$ on sides $AB$ and $AC$, respectively, such that $PQ\mid\mid BC$. Let $X$ and $Y$ be intersection points of lines $BQ$ and $CP$ with circumcircle $k$ of triangle $APQ$, and $D$ and $E$ intersection points of lines $AX$ and $AY$ with side $BC$. If $2\cdot DE=BC$, prove that circle $k$ contains intersection point of angle bisector of $\angle BAC$ with $BC$

Given triangle $\triangle ABC$ and let $D,E,F$ be the foot of angle bisectors of $A,B,C$ ,respectively. $M,N$ lie on $EF$ such that $AM=AN$. Let $H$ be the foot of $A$-altitude on $BC$. Points $K,L$ lie on $EF$ such that triangles $\triangle AKL, \triangle HMN$ are correspondingly similiar (with the given order of vertices) such that $AK \not\parallel HM$ and $AK \not\parallel HN$. Show that: $DK=DL$

Triangle $WSE$ has side lengths $WS=13$, $SE=15$, and $WE=14$. Points $J$ and $H$ lie on $\overline{SE}$ such that $SJ=JH=HE=5$. Let the angle bisector of $\angle{WES}$ intersect $\overline{WH}$ and $\overline{WJ}$ at points $M$ and $T$, respectively. Find the area of quadrilateral $JHMT$.

Given triangle $ABC, BC$ is extended beyond $B$ to the point $D$ such that $BD = BA$. The bisectors of the exterior angles at vertices $B$ and $C$ intersect at the point $M$. Prove that quadrilateral $ADMC$ is cyclic. (4)

In triangle $ABC$, the difference between angles $B$ and $C$ is equal to $90^o$, and $AL$ is the angle bisector of triangle $ABC$. The bisector of the exterior angle $A$ of the triangle $ABC$ intersects the line $BC$ at the point $F$. Prove that $AL = AF$. (Alexander Dzyunyak)

In a triangle $ABC$, the angle bisector at $A,B,C$ meet the opposite sides at $A_1,B_1,C_1$, respectively. Prove that if the quadrilateral $BA_1B_1C_1$ is cyclic, then $$\frac{AC}{AB+BC}=\frac{AB}{AC+CB}+\frac{BC}{BA+AC}.$$

Let $C$ be a point on the extension of the diameter $AB$ of a circle. A line through $C$ is tangent to the circle at point $N$. The bisector of $\angle ACN$ meets the lines $AN$ and $BN$ at $P$ and $Q$ respectively. Prove that $PN = QN$.

Let $a,b,c$ be the length sides of a given triangle and $x,y,z$ be the sides length of bisectrices, respectively. Prove the following inequality $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

Given a triangular pyramid $ABCD$. Sphere $S_1$ passing through points $A$, $B$, $C$, intersects edges $AD$, $BD$, $CD$ at points $K$, $L$, $M$, respectively; sphere $S_2$ passing through points $A$, $B$, $D$ intersects the edges $AC$, $BC$, $DC$ at points $P$, $Q$, $M$ respectively. It turned out that $KL \parallel PQ$. Prove that the bisectors of plane angles $KMQ$ and $LMP$ are the same.

Two straight lines are given on a plane, intersecting at point $O$ at an angle $a$. Let $A$, $B$ and $C $ be three points on one of the lines, located on one side of$ O$ and following in the indicated order, $M$ be an arbitrary point on another line, different from $O$, Let $\angle AMB=\gamma$, $\angle BMC = \phi$. Consider the function $F(M) = ctg \gamma + ctg \phi$ . Prove that$ F(M)$ takes the smallest value on each of the rays into which $O$ divides the second straight line. (Each has its own.) Let us denote one of these smallest values by $q$, and the other by $p$. Prove that the exprseeion $\frac{p}{q}$ is independent of choice of points $A$, $B$ and $C$. Express this relationship in terms of $a$.

Justify the following construction of the bisector of an angle with an inaccessible vertex: [img]https://cdn.artofproblemsolving.com/attachments/9/d/be4f7799d58a28cab3b4c515633b0e021c1502.png[/img] $M \in a$ and $N \in b$ are taken, the $4$ bisectors of the $4$ internal angles formed by $MN$ are traced with $a$ and $ b$. Said bisectors intersect at $P$ and $Q$, then $PQ$ is the bisector sought.

Say that a (nondegenerate) triangle is [i]funny[/i] if it satisfies the following condition: the altitude, median, and angle bisector drawn from one of the vertices divide the triangle into 4 non-overlapping triangles whose areas form (in some order) a 4-term arithmetic sequence. (One of these 4 triangles is allowed to be degenerate.) Find with proof all funny triangles.

In a triangle $ABC$ $A_0$,$B_0$ and $C_0$ are the midpoints of the sides $BC$,$CA$ and $AB$.$A_1$,$B_1$,$C_1$ are the midpoints of the broken lines $BAC,CAB,ABC$.Show that $A_0A_1,B_0B_1,C_0C_1$ are concurrent.

Let $AD$ be the bisector of triangle $ABC$. Circle $\omega$ passes through the vertex $A$ and touches the side $BC$ at point $D$. This circle intersects the sides $AC$ and $AB$ for the second time at points $M$ and $N$ respectively. Lines $BM$ and $CN$ intersect the circle for the second time $\omega$ at points $P$ and $Q$, respectively. Lines $AP$ and $AQ$ intersect side $BC$ at points $K$ and $L$, respectively. Prove that $KL=\frac12 BC$

In a triangle $ABC$, let $AA', BB'$ and $CC'$ be the bisectors. Suppose $A'B' \cap CC' =P$ and $A'C' \cap BB'= Q$. Prove that $\angle PAC = \angle QAB$.