Let $\vartriangle ABC$ be a scalene triangle with $AB < BC < CA$. Let $D$ be the projection of $A$ onto the angle bisector of $\angle ABC$, and let $E$ be the projection of $A$ onto the angle bisector of $\angle ACB$. The line $DE$ cuts sides $AB$ and AC at $M$ and $N$, respectively. Prove that $$\frac{AB+AC}{BC} =\frac{DE}{MN} + 1$$

Suppose $ABCD$ is a square piece of cardboard with side length $a$. On a plane are two parallel lines $\ell_1$ and $\ell_2$, which are also $a$ units apart. The square $ABCD$ is placed on the plane so that sides $AB$ and $AD$ intersect $\ell_1$ at $E$ and $F$ respectively. Also, sides $CB$ and $CD$ intersect $\ell_2$ at $G$ and $H$ respectively. Let the perimeters of $\triangle AEF$ and $\triangle CGH$ be $m_1$ and $m_2$ respectively. Prove that no matter how the square was placed, $m_1+m_2$ remains constant.

In a non isosceles triangle $ ABC$ let $ w$ be the angle bisector of the exterior angle at $ C$. Let $ D$ be the point of intersection of $ w$ with the extension of $ AB$. Let $ k_A$ be the circumcircle of the triangle $ ADC$ and analogy $ k_B$ the circumcircle of the triangle $ BDC$. Let $ t_A$ be the tangent line to $ k_A$ in A and $ t_B$ the tangent line to $ k_B$ in B. Let $ P$ be the point of intersection of $ t_A$ and $ t_B$. Given are the points $ A$ and $ B$. Determine the set of points $ P\equal{}P(C )$ over all points $ C$, so that $ ABC$ is a non isosceles, acute-angled triangle.

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 the acute triangle $ABC$, the midpoint of side $BC$ is called $M$. Point $X$ lies on the angle bisector of $\angle AMB$ such that $\angle BXM = 90^o$. Point $Y$ lies on the angle bisector of $\angle AMC$ such that $\angle CYM = 90^o$. Line segments $AM$ and $XY$ intersect in point $Z$. Prove that $Z$ is the midpoint of $XY$ . [asy] import geometry; unitsize (1.2 cm); pair A, B, C, M, X, Y, Z; A = (0,0); B = (4,1.5); C = (0.5,3); M = (B + C)/2; X = extension(M, incenter(A,B,M), B, B + rotate(90)*(incenter(A,B,M) - M)); Y = extension(M, incenter(A,C,M), C, C + rotate(90)*(incenter(A,C,M) - M)); Z = extension(A,M,X,Y); draw(A--B--C--cycle); draw(A--M); draw(M--interp(M,X,2)); draw(M--interp(M,Y,2)); draw(B--X, dotted); draw(C--Y, dotted); draw(X--Y); dot("$A$", A, SW); dot("$B$", B, E); dot("$C$", C, N); dot("$M$", M, NE); dot("$X$", X, NW); dot("$Y$", Y, NE); dot("$Z$", Z, S); [/asy]

Let $A_{1}$, $B_{1}$, $C_{1}$ be the feet of the altitudes of an acute-angled triangle $ABC$ issuing from the vertices $A$, $B$, $C$, respectively. Let $K$ and $M$ be points on the segments $A_{1}C_{1}$ and $B_{1}C_{1}$, respectively, such that $\measuredangle KAM = \measuredangle A_{1}AC$. Prove that the line $AK$ is the angle bisector of the angle $C_{1}KM$.

In a triangle $ABC$, points $H,M,L$ are the feet of the altitude from $C$, the median from $A$, and the angle bisector from $B$, respectively. Show that if triangle $HML$ is equilateral, then so is triangle $ABC$.

We are given a right-angled triangle $MNP$ with right angle in $P$. Let $k_M$ be the circle with center $M$ and radius $MP$, and let $k_N$ be the circle with center $N$ and radius $NP$. Let $A$ and $B$ be the common points of $k_M$ and the line $MN$, and let $C$ and $D$ be the common points of $k_N$ and the line $MN$ with with $C$ between $A$ and $B$. Prove that the line $PC$ bisects the angle $\angle APB$.

Let $ABC$ be an equilateral triangle whose angle bisectors of $B$ and $C$ intersect at $D$. Perpendicular bisectors of $BD$ and $CD$ intersect $BC$ at points $E$ and $Z$ respectively. a) Prove that $BE=EZ=ZC$. b) Find the ratio of the areas of the triangles $BDE$ to $ABC$

Let $ABC$ be a triangle. On the extensions of sides $AB$ and $CB$ towards $B$, points $C_1$ and $A_1$ are taken, respectively, so that $AC = A_1C = AC_1$. Prove that circumscribed circles of triangles $ABA_1$ and $CBC_1$ intersect on the bisector of angle $B$.

Consider triangle $ABC$ with $\angle A=2\angle B$. The angle bisectors from $A$ and $C$ intersect at $D$, and the angle bisector from $C$ intersects $\overline{AB}$ at $E$. If $\dfrac{DE}{DC}=\dfrac13$, compute $\dfrac{AB}{AC}$.

Let $ABC$ be an acute triangle and let $\Gamma$ be its circumcircle. The bisector of $\angle{A}$ intersects $BC$ at $D$, $\Gamma$ at $K$ (different from $A$), and the line through $B$ tangent to $\Gamma$ at $X$. Show that $K$ is the midpoint of $AX$ if and only if $\frac{AD}{DC}=\sqrt{2}$.

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 $ABC$ be a triangle with $\angle A = 60^\circ$, and $AA', BB', CC'$ be its internal angle bisectors. Prove that $\angle B'A'C' \le 60^\circ$.

$A, B, C$ are adjacent vertices of a regular $2n$-gon and $D$ is the vertex opposite to $B$ (so that $BD$ passes through the center of the $2n$-gon). $X$ is a point on the side $AB$ and $Y$ is a point on the side $BC$ so that $XDY = \frac{\pi}{2n}$. Show that $DY$ bisects $\angle XYC$.

Given a square $ABCD$ whose side length is $1$, $P$ and $Q$ are points on the sides $AB$ and $AD$. If the perimeter of $APQ$ is $2$ find the angle $PCQ$.

Let $ABC$ be an acute triangle with $| AB | > | AC |$. Let $D$ be the foot of the altitude from $A$ to $BC$, let $K$ be the intersection of $AD$ with the internal bisector of angle $B$, Let $M$ be the foot of the perpendicular from $B$ to $CK$ (it could be in the extension of segment $CK$) and$ N$ the intersection of $BM$ and $AK$ (it could be in the extension of the segments). Let $T$ be the intersection of$ AC$ with the line that passes through $N$ and parallel to $DM$. Prove that $BM$ is the internal bisector of the angle $\angle TBC$

Let $D$ be a point on side $[BC]$ of triangle $ABC$ such that $[AD]$ is an angle bisector, $|BD|=4$, and $|DC|=3$. Let $E$ be a point on side $[AB]$ and different than $A$ such that $m(\widehat{BED})=m(\widehat{DEC})$. If the perpendicular bisector of segment $[AE]$ meets the line $BC$ at $M$, what is $|CM|$? $ \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 7 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ \text { None of above} $

Let $ABC$ be an isosceles triangle with $AB = AC$, $P$ be the midpoint of the minor arc $AB$ of its circumcircle, and $Q$ be the midpoint of $AC$. A circumcircle of triangle $APQ$ centered at $O$ meets $AB$ for the second time at point $K$. Prove that lines $PO$ and $KQ$ meet on the bisector of angle $ABC$.

In quadrilateral $ABCD$, $\angle B = \angle D = 90$ and $AC = BC + DC$. Point $P$ of ray $BD$ is such that $BP = AD$. Prove that line $CP$ is parallel to the bisector of angle $ABD$. [i](Proposed by A.Trigub)[/i]

In $\triangle ABC$, $M$ is the midpoint of side $BC$, $AN$ bisects $\angle BAC$, $BN\perp AN$ and $\theta$ is the measure of $\angle BAC$. If sides $AB$ and $AC$ have lengths $14$ and $19$, respectively, then length $MN$ equals [asy] size(230); defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, A=14*dir(36), C=intersectionpoint(B--(9001,0), Circle(A,19)), M=midpoint(B--C), b=A+14*dir(A--C), N=foot(A, B, b); draw(N--B--A--N--M--C--A^^B--M); markscalefactor=0.1; draw(rightanglemark(B,N,A)); pair point=N; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$M$", M, S); label("$N$", N, dir(30)); label("$19$", (A+C)/2, dir(A--C)*dir(90)); label("$14$", (A+B)/2, dir(A--B)*dir(270)); [/asy] $\displaystyle \text{(A)} \ 2 \qquad \text{(B)} \ \frac{5}{2} \qquad \text{(C)} \ \frac{5}{2} - \sin \theta \qquad \text{(D)} \ \frac{5}{2} - \frac{1}{2} \sin \theta \qquad \text{(E)} \ \frac{5}{2} - \frac{1}{2} \sin \left(\frac{1}{2} \theta\right)$

The incircle of triangle $ABC$, is tangent to sides $BC,CA$ and $AB$ in $D,E$ and $F$ respectively. The reflection of $F$ with respect to $B$ and the reflection of $E$ with respect to $C$ are $T$ and $S$ respectively. Prove that the incenter of triangle $AST$ is inside or on the incircle of triangle $ABC$. [i]Proposed by Mehdi E'tesami Fard[/i]

Let $\Gamma_1$ and $\Gamma_2$ be two circles that tangents each other at point $N$, with $\Gamma_2$ located inside $\Gamma_1$. Let $A, B, C$ be distinct points on $\Gamma_1$ such that $AB$ and $AC$ tangents $\Gamma_2$ at $D$ and $E$, respectively. Line $ND$ cuts $\Gamma_1$ again at $K$, and line $CK$ intersects line $DE$ at $I$. (i) Prove that $CK$ is the angle bisector of $\angle ACB$. (ii) Prove that $IECN$ and $IBDN$ are cyclic quadrilaterals.

Let be given points $A,B$ on a circle and let $P$ be a variable point on that circle. Let point $M$ be determined by $P$ as the point that is either on segment $PA$ with $AM=MP+PB$ or on segment $PB$ with $AP+MP=PB$. Find the locus of points $M$.

Let $AG, CH$ be the angle bisectors of a triangle $ABC$. It is known that one of the intersections of the circles of triangles $ABG$ and $ACH$ lies on the side $BC$. Prove that the angle $BAC$ is $60 ^o$