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 $BC$

Let $T$ be a non-isosceles triangle and $n \ge 4$ an integer . Prove that you can divide $T$ in $n$ triangles and draw in each of them an inner bisector so that those $n$ bisectors are parallel.

Let $AL$ and $BK$ be angle bisectors in the non-isosceles triangle $ABC$ ($L$ lies on the side $BC$, $K$ lies on the side $AC$). The perpendicular bisector of $BK$ intersects the line $AL$ at point $M$. Point $N$ lies on the line $BK$ such that $LN$ is parallel to $MK$. Prove that $LN = NA$.

A line $r$ contains the points $A$, $B$, $C$, $D$ in that order. Let $P$ be a point not in $r$ such that $\angle{APB} = \angle{CPD}$. Prove that the angle bisector of $\angle{APD}$ intersects the line $r$ at a point $G$ such that: $\frac{1}{GA} + \frac{1}{GC} = \frac{1}{GB} + \frac{1}{GD}$

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 $\angle APQ$.

The bisectors $AA_1$ and $CC_1$ of the right triangle $ABC$ ($\angle B = 90^o$) intersect at point $I$. The line passing through the point $C_1$ and perpendicular on the line $AA_1$ intersects the line that passes through $A_1$ and is perpendicular on $CC_1$, at the point $K$. Prove that the midpoint of the segment $KI$ lies on segment $AC$.

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)

Let $ABC$ be a triangle with angles $\angle A = 80^\circ, \angle B = 70^\circ, \angle C = 30^\circ$. Let $P$ be a point on the bisector of $\angle BAC$ satisfying $\angle BPC =130^\circ$. Let $PX, PY, PZ$ be the perpendiculars drawn from $P$ to the sides $BC, AC, AB$, respectively. Prove that the following equation with segment lengths is satisfied $$AY^3+BZ^3+CX^3=AZ^3+BX^3+CY^3.$$

For a $\triangle ABC$ , let $A_1$ be the symmetric point of the intersection of angle bisector of $\angle BAC$ and $BC$ , where center of the symmetry is the midpoint of side $BC$, In the same way we define $B_1 $ ( on $AC$ ) and $C_1$ (on $AB$). Intersection of circumcircle of $\triangle A_1B_1C_1$ and line $AB$ is the set $\{Z,C_1 \}$, with $BC$ is the set $\{X,A_1\}$ and with $CA$ is the set $\{Y,B_1\}$. If the perpendicular lines from $X,Y,Z$ on $BC,CA$ and $ AB$ , respectively are concurrent , prove that $\triangle ABC$ is isosceles.

$ABC$ is an acute-angled triangle. Let $D$ be the point on $BC$ such that $AD$ is the bisector of $\angle A$. Let $E, F$ be the feet of perpendiculars from $D$ to $AC,AB$ respectively. Suppose the lines $BE$ and $CF$ meet at $H$. The circumcircle of triangle $AFH$ meets $BE$ at $G$ (apart from $H$). Prove that the triangle constructed from $BG$, $GE$ and $BF$ is right-angled.

Angle bisector from vertex $A$ of acute triangle $ABC$ intersects side $BC$ in point $D$, and circumcircle of $ABC$ in point $E$ (different from $A$). Let $F$ and $G$ be foots of perpendiculars from point $D$ to sides $AB$ and $AC$. Prove that area of quadrilateral $AEFG$ is equal to the area of triangle $ABC$

The triangle $ABC$ is drawn on the board such that $AB + AC = 2BC$. The bisectors $AL_1, BL_2, CL_3$ were drawn in this triangle, after which everything except the points $L_1, L_2, L_3$ was erased. Use a compass and a ruler to reconstruct triangle $ABC$.

Let $ ABC$ be an acute-angled triangle; $ AD$ be the bisector of $ \angle BAC$ with $ D$ on $ BC$; and $ BE$ be the altitude from $ B$ on $ AC$. Show that $ \angle CED > 45^\circ .$ [b][weightage 17/100][/b]

Two not necessarily equal non-intersecting wooden disks, one gray and one black, are glued to a plane. An infinite angle with one gray side and one black side can be moved along the plane so that the disks remain outside the angle, while the colored sides of the angle are tangent to the disks of the same color (the tangency points are not the vertices). Prove that it is possible to draw a ray in the angle, starting from the vertex of the angle and such that no matter how the angle is positioned, the ray passes through some fixed point of the plane. (Egor Bakaev, Ilya Bogdanov, Pavel Kozhevnikov, Vladimir Rastorguev) (Junior version [url=https://artofproblemsolving.com/community/c6h2094701p15140671]here[/url]) [hide=note]There was a mistake in the text of the problem 3, we publish here the correct version. The solutions were estimated according to the text published originally.[/hide]

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 $\angle BOC=\angle AOD$.

In $\Delta ABC$, $\angle ABC=120^\circ$. The internal bisector of $\angle B$ meets $AC$ at $D$. If $BD=1$, find the smallest possible value of $4BC+AB$.

Let $BL$ be angle bisector of acute triangle $ABC$.Point $K$ choosen on $BL$ such that $\measuredangle AKC-\measuredangle ABC=90º$.point $S$ lies on the extention of $BL$ from $L$ such that $\measuredangle ASC=90º$.Point $T$ is diametrically opposite the point $K$ on the circumcircle of $\triangle AKC$.Prove that $ST$ passes through midpoint of arc $ABC$.(S. Berlov) [hide] :trampoline: my 100th post :trampoline: [/hide]

Let $ABC$ be a triangle, $I$ its incenter, and $D$ a point on the arc $BC$ of the circumcircle of $ABC$ not containing $A$. The bisector of the angle $\angle ADB$ intesects the segment $AB$ at $E$. The bisector of the angle $\angle CDA$ intesects the segment $AC$ at $F$. Prove that the points $E, F,I$ are collinear. (Malik Talbi)

$\triangle{ABC}$ is isosceles with $AB = AC >BC$. Let $D$ be a point in its interior such that $DA = DB+DC$. Suppose that the perpendicular bisector of $AB$ meets the external angle bisector of $\angle{ADB}$ at $P$, and let $Q$ be the intersection of the perpendicular bisector of $AC$ and the external angle bisector of $\angle{ADC}$. Prove that $B,C,P,Q$ are concyclic.

Let $ABCD$ be a square and let $\Gamma$ be the circumcircle of $ABCD$. $M$ is a point of $\Gamma$ belonging to the arc $CD$ which doesn't contain $A$. $P$ and $R$ are respectively the intersection points of $(AM)$ with $[BD]$ and $[CD]$, $Q$ and $S$ are respectively the intersection points of $(BM)$ with $[AC]$ and $[DC]$. Prove that $(PS)$ and $(QR)$ are perpendicular.

Let $ABC$ be a triangle such that $AB<AC$. The perpendicular bisector of the side $BC$ meets the side $AC$ at the point $D$, and the (interior) bisectrix of the angle $ADB$ meets the circumcircle $ABC$ at the point $E$. Prove that the (interior) bisectrix of the angle $AEB$ and the line through the incentres of the triangles $ADE$ and $BDE$ are perpendicular.

Let $\triangle ABC$ have incenter $I$ and centroid $G$. Suppose that $P_A$ is the foot of the perpendicular from $C$ to the exterior angle bisector of $B$, and $Q_A$ is the foot of the perpendicular from $B$ to the exterior angle bisector of $C$. Define $P_B$, $P_C$, $Q_B$, and $Q_C$ similarly. Show that $P_A, P_B, P_C, Q_A, Q_B,$ and $Q_C$ lie on a circle whose center is on line $IG$.

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 $\angle DFE$.

In a triangle $ABC$, let $M$ be the midpoint of $BC$ and $I$ the incenter of $ABC$. If $IM$ = $IA$, find the least possible measure of $\angle{AIM}$.

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 $AC$.