Let $ ABC$ is a triangle, let $ H$ is orthocenter of $ \triangle ABC$, let $ M$ is midpoint of $ BC$. Let $ (d)$ is a line perpendicular with $ HM$ at point $ H$. Let $ (d)$ meet $ AB, AC$ at $ E, F$ respectively. Prove that $ HE \equal{}HF$.

Given a triangle $ABC$ with side $AB$ greater than $BC$, let $M$ be the midpoint of $AC$ and $L$ be the point at which the bisector of angle $\angle B$ intersects side $AC$. The line parallel to $AB$, which intersects the bisector $BL$ at $D$, is drawn by $M$, and the line parallel to the side $BC$ that intersects the median $BM$ at $E$ is drawn by $L$. Show that $ED$ is perpendicular to $BL$.

Two circles $ \Omega_{1}$ and $ \Omega_{2}$ are externally tangent to each other at a point $ I$, and both of these circles are tangent to a third circle $ \Omega$ which encloses the two circles $ \Omega_{1}$ and $ \Omega_{2}$. The common tangent to the two circles $ \Omega_{1}$ and $ \Omega_{2}$ at the point $ I$ meets the circle $ \Omega$ at a point $ A$. One common tangent to the circles $ \Omega_{1}$ and $ \Omega_{2}$ which doesn't pass through $ I$ meets the circle $ \Omega$ at the points $ B$ and $ C$ such that the points $ A$ and $ I$ lie on the same side of the line $ BC$. Prove that the point $ I$ is the incenter of triangle $ ABC$. [i]Alternative formulation.[/i] Two circles touch externally at a point $ I$. The two circles lie inside a large circle and both touch it. The chord $ BC$ of the large circle touches both smaller circles (not at $ I$). The common tangent to the two smaller circles at the point $ I$ meets the large circle at a point $ A$, where the points $ A$ and $ I$ are on the same side of the chord $ BC$. Show that the point $ I$ is the incenter of triangle $ ABC$.

The vertices $ A$ and $ B$ of an equilateral triangle $ ABC$ lie on a circle $k$ of radius $1$, and the vertex $ C$ is in the interior of the circle $ k$. A point $ D$, different from $ B$, lies on $ k$ so that $ AD\equal{}AB$. The line $ DC$ intersects $ k$ for the second time at point $ E$. Find the length of the line segment $ CE$.

Let $AD$ and $BE$ be angle bisectors in triangle $ABC$. Let $x$, $y$ and $z$ be distances from point $M$, which lies on segment $DE$, from sides $BC$, $CA$ and $AB$, respectively. Prove that $z=x+y$

Let $AD$, $BE$ and $CF$ be the diameters of the circle circumscribed around the acute angle triangle $ABC$. Point $N$ is the midpoint of the arc $CAD$, and point $M$ is the midpoint of arc $BAD$. Prove that the lines $EN$ and $MF$ intersect at the angle bisector of $\angle BAC$. (Matvii Kurskyi)

In a triangle $ABC$ let $D$ be the intersection of the angle bisector of $\gamma$, angle at $C$, with the side $AB.$ And let $F$ be the area of the triangle $ABC.$ Prove the following inequality: \[2 \cdot \ F \cdot \left( \frac{1}{AD} -\frac{1}{BD} \right) \leq AB.\]

In triangle $ABC$, $AB=13$, $BC=14$ and $CA=15$. Segment $BC$ is split into $n+1$ congruent segments by $n$ points. Among these points are the feet of the altitude, median, and angle bisector from $A$. Find the smallest possible value of $n$. [i]Proposed by Evan Chen[/i]

In the isosceles triangle $ABC$, $ (AB = BC)$ the bisector $AD$ was drawn, and in the triangle $ABD$ the bisector $DE$ was drawn. Find the values of the angles of the triangle $ABC$, if it is known that the bisectors of the angles $ABD$ and $AED$ intersect on the line $AD$. (Fedak Ivan)

Bisectrices $AA_1$ and $BB_1$ of triangle $ABC$ meet in $I$. Segments $A_1I$ and $B_1I$ are the bases of isosceles triangles with opposite vertices $A_2$ and $B_2$ lying on line $AB$. It is known that line $CI$ bisects segment $A_2B_2$. Is it true that triangle $ABC$ is isosceles?

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 $\angle DGE + \angle DGF = 180^{\circ}$.

Let $AD$ be the diameter of a circle $\omega$ and $BC$ is a chord of $\omega$ which is perpendicular to $AD$. Let $M,N,P$ be points on the segments $AB,AC,BC$ respectively, such that $MP\parallel AC$ and $PN\parallel AB$. The line $MN$ cuts the line $PD$ in the point $Q$ and the angle bisector of $\angle MPN$ in the point $R$. Prove that the points $B,R,Q,C$ are concyclic.

Let $ABCD$ be a rectangle of centre $O$, such that $\angle DAC=60^{\circ}$. The angle bisector of $\angle DAC$ meets $DC$ at $S$. Lines $OS$ and $AD$ meet at $L$, and lines $BL$ and $AC$ meet at $M$. Prove that lines $SM$ and $CL$ are parallel.

Find all triangles $ABC$ for which $\angle ACB$ is acute and the interior angle bisector of $BC$ intersects the trisectors $(AX, (AY$ of the angle $\angle BAC$ in the points $N,P$ respectively, such that $AB=NP=2DM$, where $D$ is the foot of the altitude from $A$ on $BC$ and $M$ is the midpoint of the side $BC$.

Let $ABC$ be a scalene triangle and $\omega$ is your incircle. The sides $BC,CA$ and $AB$ are tangents to $\omega$ in $X,Y,Z$ respectively. Let $M$ be the midpoint of $BC$ and $D$ is the intersection point of $BC$ with the angle bisector of $\angle BAC$. Prove that $\angle BAX=\angle MAC$ if and only if $YZ$ passes by the midpoint of $AD$.

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 $ABC$ 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]

Show that in a right-angled triangle the bisector of the right angle divides into equal parts the angle between the altitude and the median, drawn from the same vertex.

Let $ABCD$ be an inscribed quadrilateral, in which $\angle BAD<90$. On the rays $AB$ and $AD$ are selected points $K$ and $L$, respectively, such that$ KA = KD, LA = LB$. Let $N$ - the midpoint of $AC$.Prove that if $\angle BNC=\angle DNC $,so $\angle KNL=\angle BCD $

Let $C$ be a point on the extension of the diameter $AB$ of a circle. A line through $C$ is tangent to the circle at point $N$. The bisector of $\angle ACN$ meets the lines $AN$ and $BN$ at $P$ and $Q$ respectively. Prove that $PN = QN$.

In isosceles triangle $ABC$ ($AC=BC$), angle bisector $\angle BAC$ and altitude $CD$ from point $C$ intersect at point $O$, such that $CO=3 \cdot OD$. In which ratio does altitude from point $A$ on side $BC$ divide altitude $CD$ of triangle $ABC$

Let $ABC$ be a triangle with $\angle ABC=120^o$ and triangle bisectors $(AA_1),(BB_1),(CC_1)$, respectively. $B_1F \perp A_1C_1$, where $F\in (A_1C_1)$. Let $R,I$ and $S$ be the centers of the circles which are inscribed in triangles $C_1B_1F,C_1B_1A_1, A_1B_1F$, and $B_1S\cap A_1C_1=\{Q\}$. Show that $R,I,S,Q$ are on the same circle.

In quadrilateral $ABCD$ sides $BC$ and $AD$ are parallel. In each of the four vertices we draw an angular bisector. The angular bisectors of angles $A$ and $B$ intersect in point $P$, those of angles $B$ and $C$ intersect in point $Q$, those of angles $C$ and $D$ intersect in point $R$, and those of angles $D$ and $A$ intersect in point S. Suppose that $PS$ is parallel to $QR$. Prove that $|AB| =|CD|$. [asy] unitsize(1.2 cm); pair A, B, C, D, P, Q, R, S; A = (0,0); D = (3,0); B = (0.8,1.5); C = (3.2,1.5); S = extension(A, incenter(A,B,D), D, incenter(A,C,D)); Q = extension(B, incenter(A,B,C), C, C + incenter(A,B,D) - A); P = extension(A, S, B, Q); R = extension(D, S, C, Q); draw(A--D--C--B--cycle); draw(B--Q--C); draw(A--S--D); dot("$A$", A, SW); dot("$B$", B, NW); dot("$C$", C, NE); dot("$D$", D, SE); dot("$P$", P, dir(90)); dot("$Q$", Q, dir(270)); dot("$R$", R, dir(90)); dot("$S$", S, dir(90)); [/asy] Attention: the figure is not drawn to scale.

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]

Let $ABCD$ be a convex quadrilateral with $\angle{DAC}= \angle{BDC}= 36^\circ$ , $\angle{CBD}= 18^\circ$ and $\angle{BAC}= 72^\circ$. The diagonals and intersect at point $P$ . Determine the measure of $\angle{APD}$.