Given a triangle $ ABC$, let $ r$ be the external bisector of $ \angle ABC$. $ P$ and $ Q$ are the feet of the perpendiculars from $ A$ and $ C$ to $ r$. If $ CP \cap BA \equal{} M$ and $ AQ \cap BC\equal{}N$, show that $ MN$, $ r$ and $ AC$ concur.

Bisectors of angle $C$ and externalangle $A$ of trapezoid $ABCD$ with bases $BC$ and $AD$ intersect at point $M$, and the bisector of angle $B$ and external angle $D$ intersect at point $N$. Prove that the midpoint of the segment $MN$ is equidistant from the lines $AB$ and $CD$.

The bisectors of angles $A$ and $C$ of triangle $ABC$ intersect its sides at points $A_1$ and $C_1$, and the circumcircle of this triangle is at points $A_0$ and $C_0$, respectively. Lines $A_1C_1$ and $A_0C_0$ intersect at point P. Prove that the segment connecting $P$ to the center of the incircle of triangle $ABC$ is parallel to $AC$.

In trapezoid $ABCD$, the bisectors of angles $A$ and $D$ intersect at point $E$ lying on the side of $BC$. These bisectors divide the trapezoid into three triangles into which the circles are inscribed. One of these circles touches the base $AB$ at the point $K$, and two others touch the bisector $DE$ at points $M$ and $N$. Prove that $BK = MN$.

Let $S$ be a great circle with pole $P$. On any great circle through $P$, two points $A$ and $B$ are chosen equidistant from $P$. For any [i] spherical triangle [/i] $ABC$ (the sides are great circles ares), where $C$ is on $S$, prove that the great circle are $CP$ is the angle bisector of angle $C$. [b] Note. [/b] A great circle on a sphere is one whose center is the center of the sphere. A pole of the great circle $S$ is a point $P$ on the sphere such that the diameter through $P$ is perpendicular to the plane of $S$.

Two circles touch internally at point $A$. A chord $ BC $ of the larger circle is drawn tangent to the smaller one at point $ D $. Prove that $ AD $ is the bisector of angle $ BAC $.

In triangle $ABC$ the angular bisector from $A$ intersects the side $BC$ at the point $D$, and the angular bisector from $B$ intersects the side $AC$ at the point $E$. Furthermore $|AE| + |BD| = |AB|$. Prove that $\angle C = 60^o$ [img]https://1.bp.blogspot.com/-8ARqn8mLn24/XzP3P5319TI/AAAAAAAAMUQ/t71-imNuS18CSxTTLzYXpd806BlG5hXxACLcBGAsYHQ/s0/2018%2BMohr%2Bp5.png[/img]

Quadrangle $ ABCD$ is inscribed in a circle with radius 1 in such a way that the diagonal $ AC$ is a diameter of the circle, while the other diagonal $ BD$ is as long as $ AB$. The diagonals intersect at $ P$. It is known that the length of $ PC$ is $ 2/5$. How long is the side $ CD$?

Let $AC>BC$ in $\triangle ABC$ and $M$ and $N$ be the midpoints of $AC$ and $BC$ respectively. The angle bisector of $\angle B$ intersects $\overline{MN}$ at $P$. The incircle of $\triangle ABC$ has center $I$ and touches $BC$ at $Q$. The perpendiculars from $P$ and $Q$ to $MN$ and $BC$ respectively intersect at $R$. Let $S=AB\cap RN$. a) Prove that $PCQI$ is cyclic b) Express the length of the segment $BS$ with $a$, $b$, $c$ - the side lengths of $\triangle ABC$ .

Let $ABCD$ be a parallelogram ($AB < BC$). The bisector of the angle $BAD$ intersects the side $BC$ at the point K; and the bisector of the angle $ADC$ intersects the diagonal $AC$ at the point $F$. Suppose that $KD \perp BC$. Prove that $KF \perp BD$.

Given a circumcircle $(O)$ and two fixed points $B,C$ on $(O)$. $BC$ is not the diameter of $(O)$. A point $A$ varies on $(O)$ such that $ABC$ is an acute triangle. $E,F$ is the foot of the altitude from $B,C$ respectively of $ABC$. $(I)$ is a variable circumcircle going through $E$ and $F$ with center $I$. a) Assume that $(I)$ touches $BC$ at $D$. Probe that $\frac{DB}{DC}=\sqrt{\frac{\cot B}{\cot C}}$. b) Assume $(I)$ intersects $BC$ at $M$ and $N$. Let $H$ be the orthocenter and $P,Q$ be the intersections of $(I)$ and $(HBC)$. The circumcircle $(K)$ going through $P,Q$ and touches $(O)$ at $T$ ($T$ is on the same side with $A$ wrt $PQ$). Prove that the interior angle bisector of $\angle{MTN}$ passes through a fixed point.

On the circle $K$ there are given three distinct points $A,B,C$. Construct (using only a straightedge and a compass) a fourth point $D$ on $K$ such that a circle can be inscribed in the quadrilateral thus obtained.

Two circles are tangent to each other internally at a point $T$. Let the chord $AB$ of the larger circle be tangent to the smaller circle at a point $P$. Prove that the line TP bisects $\angle ATB$.

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]

$\triangle ABC$ has $AB=4$ and $AC=6$. Let point $D$ be on line $AB$ so that $A$ is between $B$ and $D$. Let the angle bisector of $\angle BAC$ intersect line $BC$ at $E$, and let the angle bisector of $\angle DAC$ intersect line $BC$ at $F$. Given that $AE=AF$, find the square of the circumcircle's radius' length.

A circle $k = (S, r)$ is given and a hexagon $AA'BB'CC'$ inscribed in it. The lengths of sides of the hexagon satisfy $AA' = A'B, BB' = B'C, CC' = C'A$. Prove that the area $P$ of triangle $ABC$ is not greater than the area $P'$ of triangle $A'B'C'$. When does $P = P'$ hold?

In the triangle $ABC$, we have that $\angle BAC$ is acute. Let $\Gamma$ be the circle that passes through $A$ and is tangent to the side $BC$ at $C$. Let $M$ be the midpoint of $BC$ and let $D$ be the other point of intersection of $\Gamma$ with $AM$. If $BD$ cuts back to$ \Gamma$ at $E$, show that $AC$ is the bisector of $\angle BAE$.

Let $ABCD$ be a parallelogram such that $AB = 35$ and $BC = 28$. Suppose that $BD \perp BC$. Let $\ell_1$ be the reflection of $AC$ across the angle bisector of $\angle BAD$, and let $\ell_2$ be the line through $B$ perpendicular to $CD$. $\ell_1$ and $\ell_2$ intersect at a point $P$. If $PD$ can be expressed in simplest form as $\frac{m}{n}$, find $m + n$.

The perpendicular bisector to side $AC$ of triangle $ABC$ intersects side $BC$ at point $M$ (see fig.). The bisector of angle $\angle AMB$ intersects the circumcircle of triangle $ABC$ at point $K$. Prove that the line passing through the centers of the inscribed circles triangles $AKM$ and $BKM$, perpendicular to the bisector of angle $\angle AKB$. [img]https://cdn.artofproblemsolving.com/attachments/b/4/b53ec7df0643a90b835f142d99c417a2a1dd45.png[/img]

Two circles touch internally at point $P$. A line tangent to one of the circles at point $A$ intersects the other circle at points $B$ and $C$. Prove that the line $ PA $ is the bisector of the angle $ BPC $.

Let $A$ and $B$ be points on the circle $k$ with center $O$, so that $AB> AO$. Let $C$ be the intersection of the bisectors of $\angle OAB$ and $k$, different from $A$. Let $D$ be the intersection of the straight line $AB$ with the circumcircle of the triangle $OBC$, different from $B$. Show that $AD = AO$ .

In an acute triangle $ABC$ , the bisector of angle $\angle A$ intersects the circumscribed circle of the triangle $ABC$ at the point $W$. From point $W$ , a parallel is drawn to the side $AB$, which intersects this circle at the point $F \ne W$. Describe the construction of the triangle $ABC$, if given are the segments $FA$ , $FW$ and $\angle FAC$. (Andrey Mostovy)

In the triangle $ABC$ the angle bisectors $AD$ and $BE$ are drawn. Prove that $\angle ACB = 60 {} ^ \circ$ if and only if $AE + BD = AB$. (Hilko Danilo)

Let $\Gamma$ be the circumcircle of an acute triangle $ABC$. The perpendicular line to $AB$ passing through $C$ cuts $AB$ in $D$ and $\Gamma$ again in $E$. The bisector of the angle $C$ cuts $AB$ in $F$ and $\Gamma$ again in $G$. The line $GD$ meets $\Gamma$ again at $H$ and the line $HF$ meets $\Gamma$ again at $I$. Prove that $AI = EB$.

In $\triangle ABC$, $AB = 40$, $BC = 60$, and $CA = 50$. The angle bisector of $\angle A$ intersects the circumcircle of $\triangle ABC$ at $A$ and $P$. Find $BP$. [i]Proposed by Eugene Chen[/i]