Let $ABC$ be an acute-angled triangle with $\angle ACB = 60^o$ , and its heights $AD$ and $BE$ intersect at point $H$. Prove that the circumcenter of triangle $ABC$ lies on a line bisecting the angles $AHE$ and $BHD$.

In a non-equilateral acute-angled triangle $ABC$ with $\angle C = 60^\circ$, $U$ is the circumcenter, $H$ the orthocenter and $D$ the intersection of $AH$ and $BC$. Prove that the Euler line $HU$ bisects the angle $BHD$.

Given an acute triangle $ ABC$. The incircle of triangle $ ABC$ touches $ BC,CA,AB$ respectively at $ D,E,F$. The angle bisector of $ \angle A$ cuts $ DE$ and $ DF$ respectively at $ K$ and $ L$. Suppose $ AA_1$ is one of the altitudes of triangle $ ABC$, and $ M$ be the midpoint of $ BC$. (a) Prove that $ BK$ and $ CL$ are perpendicular with the angle bisector of $ \angle BAC$. (b) Show that $ A_1KML$ is a cyclic quadrilateral.

Let $ABC$ be an acute non isosceles triangle. The angle bisector of angle $A$ meets again the circumcircle of the triangle $ABC$ in $D$. Let $O$ be the circumcenter of the triangle $ABC$. The angle bisectors of $\angle AOB$, and $\angle AOC$ meet the circle $\gamma$ of diameter $AD$ in $P$ and $Q$ respectively. The line $PQ$ meets the perpendicular bisector of $AD$ in $R$. Prove that $AR // BC$.

The sides of triangle $ABC$ are $a = | BC |, b = | CA |$ and $c = | AB |$. Points $D, E$ and $F$ are the points on the sides $BC, CA$ and $AB$ such that $AD, BE$ and $CF$ are the angle bisectors of the triangle $ABC$. Determine the lengths of the segments $AD, BE$, and $CF$ in terms of $a, b$, and $c$.

Does a parallelogram exist such that all pairwise meets of bisectors of its angles are situated outside it?

In a trapezoid $ABCD$ with bases $AD$ and $BC$, the bisector of the angle $\angle DAB$ intersects the bisectors of the angles $\angle ABC$ and $\angle CDA$ at the points $P$ and $S$, respectively, and the bisector of the angle $\angle BCD$ intersects the bisectors of the angles $\angle ABC$ and $\angle CDA$ at the points $Q$ and $R$, respectively. Prove that if $PS\parallel RQ$, then $AB = CD$.

Let $\triangle ABC$ be a triangle with $AB<AC$. Let the angle bisector of $\angle BAC$ meet $BC$ at $D$, and let $M$ be the midpoint of $\overline{BC}$. Let $P$ be the foot of the perpendicular from $B$ to $\overline{AD}$. Extend $\overline{BP}$ to meet $\overline{AM}$ at $Q$. Show that $\overline{DQ}$ is parallel to $\overline{AB}$.

Given a convex pentagon $ABCDE$ such that $AE$ is parallel to $CD$ and $AB=BC$. Angle bisectors of angles $A$ and $C$ intersect at $K$. Prove that $BK$ and $AE$ are parallel.

Let $F$ be an intersection point of altitude $CD$ and internal angle bisector $AE$ of right angled triangle $ABC$, $\angle ACB = 90^{\circ}$. Let $G$ be an intersection point of lines $ED$ and $BF$. Prove that area of quadrilateral $CEFG$ is equal to area of triangle $BDG$

Let $ABCD$ be a convex quadrilateral and let $P$ be the point of intersection of $AC$ and $BD$. Suppose that $AC+AD=BC+BD$. Prove that the internal angle bisectors of $\angle ACB$, $\angle ADB$ and $\angle APB$ meet at a common point.

Let $\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $ABC$ with $AB > BC$. The angle bisector of $\angle ABC$ intersects $\Omega$ at $M \ne B$. Let $\Gamma$ be the circle with diameter $BM$. The angle bisectors of $\angle AOB$ and $\angle BOC$ intersect $\Gamma$ at points $P$ and $Q,$ respectively. The point $R$ is chosen on the line $P Q$ so that $BR = MR$. Prove that $BR\parallel AC$. (Here we always assume that an angle bisector is a ray.) [i]Proposed by Sergey Berlov, Russia[/i]

In triangle $ABC$, $\angle B=2\angle C$ and $\angle A>90^\circ$. Let $D$ be the point on the line $AB$ such that $CD$ is perpendicular to $AC$, and let $M$ be the midpoint of $BC$. Prove that $\angle AMB=\angle DMC$.

Let $AL$ be the bisector of triangle $ABC$. Circle $\omega_1$ is circumscribed around triangle $ABL$. Tangent to $\omega_1$ at point $B$ intersects the extension of $AL$ at point $K$. The circle $\omega_2$ circumscribed around the triangle $CKL$ intersects $\omega_1$ a second time at point $Q$, with $Q$ lying on the side $AC$. Find the value of the angle $ABC$. (Vladislav Radomsky)

The point $M$ is the center of a regular pentagon $ABCDE$. The point $P$ is an inner point of the line segment $[DM]$. The circumscribed circle of triangle $\vartriangle ABP$ intersects the side $[AE]$ at point $Q$ (different from $A$). The perpendicular from $P$ on $CD$ intersects the side $[AE] $ at point $S$. Prove that $PS$ is the bisector of $\angle APQ$.

$P,Q$ are two fixed points on a circle centered at $O$, and $M$ is an interior point of the circle that differs from $O$. $M,P,Q,O$ are concyclic. Prove that the bisector of $\angle PMQ$ is perpendicular to line $OM$.

Let $ABCD$ be a cyclic quadrilatetral inscribed in a circle $k$. Let $M$ and $N$ be the midpoints of the arcs $AB$ and $CD$ which do not contain $C$ and $A$ respectively. If $MN$ meets side $AB$ at $P$, then show that $\frac{AP}{BP}=\frac{AC+AD}{BC+BD}$

In the quadrilateral $ABCD$ the condition $AD = AB + CD$ is fulfilled. The bisectors of the angles $BAD$ and $ADC$ intersect at the point $P $, as shown in Fig. Prove that $BP = CP$. [img]https://cdn.artofproblemsolving.com/attachments/3/1/67268635aaef9c6dc3363b00453b327cbc01f3.png[/img] (Maria Rozhkova)

Given triangle $ABC$, $D$ is the foot of the external angle bisector of $A$, $I$ its incenter and $I_a$ its $A$-excenter. Perpendicular from $I$ to $DI_a$ intersects the circumcircle of triangle in $A'$. Define $B'$ and $C'$ similarly. Prove that $AA',BB'$ and $CC'$ are concurrent. [i]proposed by Amirhossein Zabeti[/i]

For an isosceles triangle $ABC$ where $AB=AC$, it is possible to construct, using only compass and straightedge, an isosceles triangle $PQR$ where $PQ=PR$ such that triangle $PQR$ is similar to triangle $ABC$, point $P$ is in the interior of line segment $AC$, point $Q$ is in the interior of line segment $AB$, and point $R$ is in the interior of line segment $BC$. Describe one method of performing such a construction. Your method should work on every isosceles triangle $ABC$, except that you may choose an upper limit or lower limit on the size of angle $BAC$. [asy] defaultpen(linewidth(0.7)); pair a= (79,164),b=(19,22),c=(138,22),p=(109,91),q=(38,67),r=(78,22); pair point = ((p.x+q.x+r.x)/3,(p.y+q.y+r.y)/3); draw(a--b--c--cycle); draw(p--q--r--cycle); label("$A$",a,dir(point--a)); label("$B$",b,dir(point--b)); label("$C$",c,dir(point--c)); label("$P$",p,dir(point--p)); label("$Q$",q,dir(point--q)); label("$R$",r,dir(point--r));[/asy]

Let the points $O$ and $H$ be the circumcenter and orthocenter of an acute angled triangle $ABC.$ Let $D$ be the midpoint of $BC.$ Let $E$ be the point on the angle bisector of $\angle BAC$ such that $AE\perp HE.$ Let $F$ be the point such that $AEHF$ is a rectangle. Prove that $D,E,F$ are collinear.

Let the bisector of the outside angle of $A$ of triangle $ABC$ and the circumcircle of triangle $ABC$ meet at point $P$. The circle passing through points $A$ and $P$ intersects segments $BP$ and $CP$ at points $E$ and $F$ respectively. Let $AD$ is the angle bisector of triangle $ABC$. Prove that $\angle PED = \angle PFD$. [img]https://cdn.artofproblemsolving.com/attachments/0/3/0638429a220f07227703a682479ed150302aae.png[/img]

Let the points $O$ and $H$ be the circumcenter and orthocenter of an acute angled triangle $ABC.$ Let $D$ be the midpoint of $BC.$ Let $E$ be the point on the angle bisector of $\angle BAC$ such that $AE\perp HE.$ Let $F$ be the point such that $AEHF$ is a rectangle. Prove that $D,E,F$ are collinear.

In an acute-angled triangle $ABC$, point $I$ is the incenter, $H$ is the orthocenter, $O$ is the center of the circumscribed circle, $T$ and $K$ are the touchpoints of the $A$-excircle and incircle with side $BC$ respectively. It turned out that the segment $TI$ is passing through the point $O$. Prove that $HK$ is the angle bisector of $\angle BHC$. (Matvii Kurskyi)

Let $D$, $E$, $F$ be the midpoints of the arcs $BC$, $CA$, $AB$ on the circumcircle of a triangle $ABC$ not containing the points $A$, $B$, $C$, respectively. Let the line $DE$ meets $BC$ and $CA$ at $G$ and $H$, and let $M$ be the midpoint of the segment $GH$. Let the line $FD$ meet $BC$ and $AB$ at $K$ and $J$, and let $N$ be the midpoint of the segment $KJ$. a) Find the angles of triangle $DMN$; b) Prove that if $P$ is the point of intersection of the lines $AD$ and $EF$, then the circumcenter of triangle $DMN$ lies on the circumcircle of triangle $PMN$.