Through the point $ K $ lying outside the circle $ \omega $, the tangents are drawn $ KB $ and $ KD $ to this circle ($ B $ and $ D $ are tangency points) and a line intersecting a circle at points $ A $ and $ C $. The bisector of angle $ ABC $ intersects the segment $ AC $ at the point $ E $ and circle $ \omega $ at $ F $. Prove that $ \angle FDE = 90^\circ $.

Consider $ n$ points in a plane which are vertices of a convex polygon. Prove that the set of the lengths of the sides and the diagonals of the polygon has at least $ \lfloor n/2\rfloor$ elements.

Let $ABCD$ be a cyclic quadrangle whose midpoints of diagonals $AC$ and $BD$ are $F$ and $G$, respectively. a) Prove the following implication: if the bisectors of angles at $B$ and $D$ of the quadrangle intersect at diagonal $AC$ then $\frac14 \cdot |AC| \cdot |BD| = | AG| \cdot |BF| \cdot |CG| \cdot |DF|$. b) Does the converse implication also always hold?

Points $X$ and $Y$ are the midpoints of arcs $AB$ and $BC$ of the circumscribed circle of triangle $ABC$. Point $T$ lies on side $AC$. It turned out that the bisectors of the angles $ATB$ and $BTC$ pass through points $X$ and $Y$ respectively. What angle $B$ can be in triangle $ABC$?

Points $A_1,~ B_1,~ C_1$ lie on the sides $BC,~ AC$ and $AB$ of a triangle $ABC$, respectively, such that $AB_1 -AC_1 = CA_1 -CB_1 = BC_1 -BA_1$. Let $I_A,~ I_B,~ I_C$ be the incenters of triangles $AB_1C_1,~ A_1BC_1$ and $A_1B_1C$ respectively. Prove that the circumcenter of triangle $I_AI_BI_C$, is the incenter of triangle $ABC$.

Let $ABC$ be a triangle in which $\measuredangle{A}=135^{\circ}$. The perpendicular to the line $AB$ erected at $A$ intersects the side $BC$ at $D$, and the angle bisector of $\angle B$ intersects the side $AC$ at $E$. Find the measure of $\measuredangle{BED}$.

Let $ABCD$ be a parallelogram, such that the point $M$ is the midpoint of the side $CD$ and lies on the bisector of the angle $\angle BAD$. Prove that $\angle AMB = 90^o$.

Let $ABC$ be a triangle. Let $A_1$, $A_2$ be points on $AB$ and $AC$ respectively such that $A_1A_2 \parallel BC$ and the circumcircle of $\triangle AA_1A_2$ is tangent to $BC$ at $A_3$. Define $B_3$, $C_3$ similarly. Prove that $AA_3$, $BB_3$, and $CC_3$ are concurrent. [i]Carl Lian.[/i]

Let $ABCD$ be a convex quadrilateral and let $P$ and $Q$ be the points on the sides $AD$ and $BC$ respectively such that $\frac{AP}{PD}=\frac{BQ}{QC}=\frac{AB}{CD}$. Prove that the line $PQ$ forms equal angles with the lines $AB$ and $CD$.

The diagonals $ AC$ and $ BD$ of a convex quadrilateral $ ABCD$ intersect at point $ M$. The bisector of $ \angle ACD$ meets the ray $ BA$ at $ K$. Given that $ MA \cdot MC \plus{}MA \cdot CD \equal{} MB \cdot MD$, prove that $ \angle BKC \equal{} \angle CDB$.

Let $ABCDEF$ be a convex hexagon such that $\angle A = \angle C = \angle E$ and $\angle B = \angle D = \angle F$ and the (interior) angle bisectors of $\angle A, ~\angle C,$ and $\angle E$ are concurrent. Prove that the (interior) angle bisectors of $\angle B, ~\angle D, $ and $\angle F$ must also be concurrent. [i]Note that $\angle A = \angle FAB$. The other interior angles of the hexagon are similarly described.[/i]

A convex quadrangle $ABCD$ is given. Let $I$ and $J$ be the circles of circles inscribed in the triangles $ABC$ and $ADC$, respectively, and $I_a$ and $J_a$ are the centers of the excircles circles of triangles $ABC$ and $ADC$, respectively (inscribed in the angles $BAC$ and $DAC$, respectively). Prove that the intersection point $K$ of the lines $IJ_a$ and $JI_a$ lies on the bisector of the angle $BCD$.

In the triangle $ABC, \angle BAC$ is acute, the angle bisector of $\angle BAC$ meets $BC$ at $D, K$ is the foot of the perpendicular from $B$ to $AC$, and $\angle ADB = 45^o$. Point $P$ lies between $K$ and $C$ such that $\angle KDP = 30^o$. Point $Q$ lies on the ray $DP$ such that $DQ = DK$. The perpendicular at $P$ to $AC$ meets $KD$ at $L$. Prove that $PL^2 = DQ \cdot PQ$.

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 $AC$.

Given a triangle $ABC$. The points $A$, $B$, $C$ divide the circumcircle $\Omega$ of the triangle $ABC$ into three arcs $BC$, $CA$, $AB$. Let $X$ be a variable point on the arc $AB$, and let $O_{1}$ and $O_{2}$ be the incenters of the triangles $CAX$ and $CBX$. Prove that the circumcircle of the triangle $XO_{1}O_{2}$ intersects the circle $\Omega$ in a fixed point.

On the side $BC$ of the triangle $ABC$ a point $D$ different from $B$ and $C$ is chosen so that the bisectors of the angles $ACB$ and $ADB$ intersect on the side $AB$. Let $D'$ be the symmetrical point to $D$ with respect to the line $AB$. Prove that the points $C, A$ and $D'$ are on the same line.

Given triangle $ ABC$ of area 1. Let $ BM$ be the perpendicular from $ B$ to the bisector of angle $ C$. Determine the area of triangle $ AMC$.

Points $A_1$, $B_1$, $C_1$ are midpoints of sides $BC$, $AC$, $AB$ of triangle $ABC$. On midlines $C_1B_1$ and $A_1B_1$ points $E$ and $F$ are chosen such that $BE$ is the angle bisector of $AEB_1$ and $BF$ is the angle bisector of $CFB_1$. Prove that bisectors of angles $ABC$ and $FBE$ coincide. [I]Proposed by F. Baharev[/i]

In triangle $ABC$ we have $|AB| \ne |AC|$. The bisectors of $\angle ABC$ and $\angle ACB$ meet $AC$ and $AB$ at $E$ and $F$, respectively, and intersect at I. If $|EI| = |FI|$ find the measure of $\angle BAC$.

Given a triangle $ABC$ inscribed in circle $(O)$ and let $P$ be a point on the interior angle bisector of $BAC$. $PB$, $PC$ cut $CA$, $AB$ at $E,F$ respectively. Let $EF$ meet $(O)$ at $M,N$. The line that is perpendicular to $PM$, $PN$ at $M,N$ respectively intersect $(O)$ at $S, T$ different from $M,N$. Prove that $ST \parallel BC$.

Let $ ABC$ be a triangle. The $ A$-excircle of triangle $ ABC$ has center $ O_{a}$ and touches the side $ BC$ at the point $ A_{a}$. The $ B$-excircle of triangle $ ABC$ touches its sidelines $ AB$ and $ BC$ at the points $ C_{b}$ and $ A_{b}$. The $ C$-excircle of triangle $ ABC$ touches its sidelines $ BC$ and $ CA$ at the points $ A_{c}$ and $ B_{c}$. The lines $ C_{b}A_{b}$ and $ A_{c}B_{c}$ intersect each other at some point $ X$. Prove that the quadrilateral $ AO_{a}A_{a}X$ is a parallelogram. [i]Remark.[/i] The $ A$[i]-excircle[/i] of a triangle $ ABC$ is defined as the circle which touches the segment $ BC$ and the extensions of the segments $ CA$ and $ AB$ beyound the points $ C$ and $ B$, respectively. The center of this circle is the point of intersection of the interior angle bisector of the angle $ CAB$ and the exterior angle bisectors of the angles $ ABC$ and $ BCA$. Similarly, the $ B$-excircle and the $ C$-excircle of triangle $ ABC$ are defined. [hide="Source of the problem"][i]Source of the problem:[/i] Theorem (88) in: John Sturgeon Mackay, [i]The Triangle and its Six Scribed Circles[/i], Proceedings of the Edinburgh Mathematical Society 1 (1883), pages 4-128 and drawings at the end of the volume.[/hide]

$ABC$ is a triangle such that $BC = 10$, $CA = 12$. Let $M$ be the midpoint of side $AC$. Given that $BM$ is parallel to the external bisector of $\angle A$, find area of triangle $ABC$. (Lines $AB$ and $AC$ form two angles, one of which is $\angle BAC$. The external angle bisector of $\angle A$ is the line that bisects the other angle.

In a triangle $ABC$, the external bisector of $\angle BAC$ intersects the ray $BC$ at $D$. The feet of the perpendiculars from $B$ and $C$ to line $AD$ are $E$ and $F$, respectively and the foot of the perpendicular from $D$ to $AC$ is $G$. Show that $\angle DGE + \angle DGF = 180^{\circ}$.

Given a tetrahedron $ABCD.$ The sphere $\omega$ inscribed in it touches the face $ABC$ at point $T$. Sphere $\omega' $ touches face $ABC$ at point $T'$ and extensions of faces $ABD$, $BCD$, $CAD$. Prove that the lines $AT$ and $AT'$ are symmetric wrt bisector of angle $\angle BAC$

Suppose that $ B_1$ is the midpoint of the arc $ AC$, containing $ B$, in the circumcircle of $ \triangle ABC$, and let $ I_b$ be the $ B$-excircle's center. Assume that the external angle bisector of $ \angle ABC$ intersects $ AC$ at $ B_2$. Prove that $ B_2I$ is perpendicular to $ B_1I_B$, where $ I$ is the incenter of $ \triangle ABC$.