In triangle $ABC$, the angle bisectors are $BE$ and $CF$ (where $E, F$ are on the sides of the triangle), and their intersection point is $I$. Point $N$ lies on the circumcircle of $AEF$, and the angle $\angle IAN$ is right. The circumcircle of $AEF$ meets the line $NI$ a second time at the point $L$. Show that the circumcenter of $AIL$ lies on line $BC$.

In triangle $ABC$, the angle at vertex $B$ is $120^o$. Let $A_1, B_1, C_1$ be points on the sides $BC, CA, AB$ respectively such that $AA_1, BB_1, CC_1$ are bisectors of the angles of triangle $ABC$. Determine the angle $\angle A_1B_1C_1$.

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$.

In the convex pentagon $ABCDE$ all interior angles have the same measure. Prove that the perpendicular bisector of segment $EA$, the perpendicular bisector of segment $BC$ and the angle bisector of $\angle CDE$ intersect in one point.

In the triangle $ABC$, the angle - bisector $AD$ ($D \in BC$) and the median $BE$ ($E \in AC$) intersect at point $P$. Lines $AB$ and $CP$ intesect at point $F$. The parallel through $B$ to $CF$ intersects $DF$ at point $M$. Prove that $DM = BF$

Suppose $n>2$ and let $A_1,\dots,A_n$ be points on the plane such that no three are collinear. [b](a)[/b] Suppose $M_1,\dots,M_n$ be points on segments $A_1A_2,A_2A_3,\dots ,A_nA_1$ respectively. Prove that if $B_1,\dots,B_n$ are points in triangles $M_2A_2M_1,M_3A_3M_2,\dots ,M_1A_1M_n$ respectively then \[|B_1B_2|+|B_2B_3|+\dots+|B_nB_1| \leq |A_1A_2|+|A_2A_3|+\dots+|A_nA_1|\] Where $|XY|$ means the length of line segment between $X$ and $Y$. [b](b)[/b] If $X$, $Y$ and $Z$ are three points on the plane then by $H_{XYZ}$ we mean the half-plane that it's boundary is the exterior angle bisector of angle $\hat{XYZ}$ and doesn't contain $X$ and $Z$ ,having $Y$ crossed out. Prove that if $C_1,\dots ,C_n$ are points in ${H_{A_nA_1A_2},H_{A_1A_2A_3},\dots,H_{A_{n-1}A_nA_1}}$ then \[|A_1A_2|+|A_2A_3|+\dots +|A_nA_1| \leq |C_1C_2|+|C_2C_3|+\dots+|C_nC_1|\] Time allowed for this problem was 2 hours.

In an acute-angled triangle $ ABC,$ let $ AD,BE$ be altitudes and $ AP,BQ$ internal bisectors. Denote by $ I$ and $ O$ the incenter and the circumcentre of the triangle, respectively. Prove that the points $ D, E,$ and $ I$ are collinear if and only if the points $ P, Q,$ and $ O$ are collinear.

(F.Nilov, A.Zaslavsky) Let $ CC_0$ be a median of triangle $ ABC$; the perpendicular bisectors to $ AC$ and $ BC$ intersect $ CC_0$ in points $ A_c$, $ B_c$; $ C_1$ is the common point of $ AA_c$ and $ BB_c$. Points $ A_1$, $ B_1$ are defined similarly. Prove that circle $ A_1B_1C_1$ passes through the circumcenter of triangle $ ABC$.

Let $ABC$ be an acute triangle and $AD$ a height. The angle bissector of $\angle DAC$ intersects $DC$ at $E$. Let $F$ be a point on $AE$ such that $BF$ is perpendicular to $AE$. If $\angle BAE=45º$, find $\angle BFC$.

In a convex quadrilateral $ABCD$ we have $AB\cdot CD=BC\cdot DA$ and $2\angle A+\angle C=180^\circ$. Point $P$ lies on the circumcircle of triangle $ABD$ and is the midpoint of the arc $BD$ not containing $A$. It is known that the point $P$ lies inside the quadrilateral $ABCD$. Prove that $\angle BCA=\angle DCP$ [i]Proposed by S. Berlov[/i]

In acute-angled triangle $ABC$ with $|BC| < |BA|$, point $N$ is the midpoint of $AC$. The circle with diameter $AB$ intersects the bisector of $\angle B$ in two points: $B$ and $X$. Prove that $XN$ is parallel to $BC$. [img]https://cdn.artofproblemsolving.com/attachments/5/1/f0ae8f5df8f2cc1bb80de1ee1807dc845a87b3.png[/img]

Let $AD$ be the internal angle bisector of triangle $ABC$. The incircles of triangles $ABC$ and $ACD$ touch each other externally. Prove that $\angle ABC > 120^{\circ}$. (Recall that the incircle of a triangle is a circle inside the triangle that is tangent to its three sides.) [i]Proposed by Volodymyr Brayman (Ukraine)[/i]

In triangle $ABC$ the angle bisector $AK$ is perpendicular on the median is $CL$. Prove that in the triangle $BKL$ also one of angle bisectors are perpendicular to one of the medians.

Let D be an arbitrary point inside $\Delta ABC$. Let $\Gamma$ be the circumcircle of $\Delta BCD$. The external angle bisector of $\angle ABC$ meets $\Gamma$ again at $E$. The external angle bisector of $\angle ACB$ meets $\Gamma$ again at $F$. The line $EF$ meets the extension of $AB$ and $AC$ at $P$ and $Q$ respectively. Prove that the circumcircles of $\Delta BFP$ and $\Delta CEQ$ always pass through the same fixed point regardless of the position of $D$. (Assume all the labelled points are distinct.)

A circle with center O is inscribed in a quadrilateral ABCD and touches its non-parallel sides BC and AD at E and F respectively. The lines AO and DO meet the segment EF at K and N respectively, and the lines BK and CN meet at M. Prove that the points O,K,M and N lie on a circle.

Given a triangle $ \,ABC,\,$ let $ \,I\,$ be the center of its inscribed circle. The internal bisectors of the angles $ \,A,B,C\,$ meet the opposite sides in $ \,A^{\prime },B^{\prime },C^{\prime }\,$ respectively. Prove that \[ \frac {1}{4} < \frac {AI\cdot BI\cdot CI}{AA^{\prime }\cdot BB^{\prime }\cdot CC^{\prime }} \leq \frac {8}{27}. \]

A triangle $ABC$ is given in which $\angle BAC = 40^o$. and $\angle ABC = 20^o$. Find the length of the angle bisector drawn from the vertex $C$, if it is known that the sides $AB$ and $BC$ differ by $4$ centimeters.

Let $ ABCD $ be a convex quadrilateral with no pair of parallel sides, such that $ \angle ABC = \angle CDA $. Assume that the intersections of the pairs of neighbouring angle bisectors of $ ABCD $ form a convex quadrilateral $ EFGH $. Let $ K $ be the intersection of the diagonals of $ EFGH$. Prove that the lines $ AB $ and $ CD $ intersect on the circumcircle of the triangle $ BKD $.

Triangle $ ABC$ has $ AC\equal{}3$, $ BC\equal{}4$, and $ AB\equal{}5$. Point $ D$ is on $ \overline{AB}$, and $ \overline{CD}$ bisects the right angle. The inscribed circles of $ \triangle ADC$ and $ \triangle BCD$ have radii $ r_a$ and $ r_b$, respectively. What is $ r_a/r_b$? $ \textbf{(A)}\ \frac{1}{28}\left(10\minus{}\sqrt{2}\right) \qquad \textbf{(B)}\ \frac{3}{56}\left(10\minus{}\sqrt{2}\right) \qquad \textbf{(C)}\ \frac{1}{14}\left(10\minus{}\sqrt{2}\right) \qquad \textbf{(D)}\ \frac{5}{56}\left(10\minus{}\sqrt{2}\right) \\ \textbf{(E)}\ \frac{3}{28}\left(10\minus{}\sqrt{2}\right)$

Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$, $AB$, respectively. Show that $PQ=QR$ if and only if the bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$.

The bisectors $AA_1$ and $CC_1$ of triangle $ABC$ intersect at point $I$. The circumscribed circles of triangles $AIC_1$ and $CIA_1$ intersect the arcs $AC$ and $BC$ (not containing points $B$ and $A$ respectively) of the circumscribed circle of triangle $ABC$ at points $C_2$ and $A_2$, respectively. Prove that lines $A_1A_2$ and $C_1C_2$ intersect on the circumscribed circle of triangle $ABC$.

In the non-isosceles $\vartriangle ABC$, the interior bisectors of vertices $B$ and $C$ are drawn, which cut the sides $AC$ and $AB$ at $E$ and $F$ respectively.The line $EF$ cuts the extension of side $BC$ at $T$. In the side$ BC$ a point D is located, so that $\frac{DB}{DC} = \frac{TB}{TC}$. Prove that $AT$ is the exterior bisector of angle $A$.

A circle $\omega$ and a point $T$ outside the circle is given. Let a tangent from $T$ to $\omega$ touch $\omega$ at $A$, and take points $B,C$ lying on $\omega$ such that $T,B,C$ are colinear. The bisector of $\angle ATC$ intersects $AB$ and $AC$ at $P$ and $Q$,respectively. Prove that $PA=\sqrt{PB\cdot QC}$

Let $ABCD$ be a trapezium such that $AB\parallel CD$. Let $E$ be the intersection of diagonals $AC$ and $BD$. Suppose that $AB = BE$ and $AC = DE$. Prove that the internal angle bisector of $\angle BAC$ is perpendicular to $AD$.

Let $O$ be a point inside an acute angle with the vertex $A$ and $H, N$ be the feet of the perpendiculars drawn from $O$ onto the sides of the angle. Let point $B$ belong to the bisector of the angle, $K$ be the foot of the perpendicular from $B$ onto either side of the angle. Denote by $P,F$ the midpoints of the segments $AK,HN$ respectively. Known that $ON + OH = BK$, prove that $PF$ is perpendicular to $AB$. Ya. Konstantinovski