Given a trihedral angle Sxyz. A plane (P) not through S cuts Sx,Sy,Sz respectively at A,B,C. On the plane (P), outside triangle ABC, construct triangles DAB,EBC,FCA which are confruent to the triangles SAB,SBC,SCA respectively. Let (T) be the sphere lying inside Sxyz, but not inside the tetrahedron SABC, toucheing the planes containing the faces of SABC. Prove that (T) touches the plane (P) at the circumcenter of triangle DEF.

Let $ABCD$ be a convex quadrilateral inscibed in circle $(O)$ such that $DB = DA + DC$. The point $P$ lies on the ray $AC$ such that $AP = BC$. The point $E$ is on $(O)$ such that $BE \perp AD$. Prove that $DP$ is parallel to the angle bisector of $\angle BEC$.

Let $ABC$ be an acute triangle with $AB\neq AC$. The angle bisector of $\angle BAC$ intersects with $BC$ at a point $D$. $BE,CF$ are the altitudes of the triangle and $Ap_1,Ap_2$ are the isodynamic points of triangle $ABC$.Let the $A$-median of $ABC$ intersect $EF$ at $T$. Show that the line connecting $T$ with the nine-point center of $ABC$ is perpendicular to $BC$ if and only if $\angle Ap_1DAp_2=90^\circ$.

Let $ABC$ be a triangle with $\angle BAC = 90^\circ$. Construct the square $BDEC$ such as $A$ and the square are at opposite sides of $BC$. Let the angle bisector of $\angle BAC$ cut the sides $[BC]$ and $[DE]$ at $F$ and $G$, respectively. If $|AB|=24$ and $|AC|=10$, calculate the area of quadrilateral $BDGF$.

Let $ABC$ be a triangle and let $M$ be the midpoint of the side $AB$. Let $BD$ be the interior angle bisector of $\angle ABC$, $D\in AC$. Prove that if $MD \perp BD$ then $AB=3BC$.

Given triangle $ABC$, with $AC> BC$, and the it's circumcircle centered at $O$. Let $M$ be the point on the circumcircle of triangle $ABC$ so that $CM$ is the bisector of $\angle ACB$. Let $\Gamma$ be a circle with diameter $CM$. The bisector of $BOC$ and bisector of $AOC$ intersect $\Gamma$ at $P$ and $Q$, respectively. If $K$ is the midpoint of $CM$, prove that $P, Q, O, K$ lie at one point of the circle.

In the triangle $ABC$, $\angle A=90^\circ$, the bisector of $\angle B$ meets the altitude $AD$ at the point $E$, and the bisector of $\angle CAD$ meets the side $CD$ at $F$. The line through $F$ perpendicular to $BC$ intersects $AC$ at $G$. Prove that $B,E,G$ are collinear.

The points $ P, Q$ are given on the plane, which are the points of intersection of the angle bisector $AL$ of some triangle $ABC$ with an inscribed circle, and the point $W$ is the intersection of the angle bisector $AL$ with a circumscribed circle other than the vertex $A$. a) Find the geometric locus of the possible location of the vertex $A$ of the triangle $ABC$. b) Find the geometric locus of the possible location of the vertex $B$ of the triangle $ABC$.

Let $ \alpha$ and $ \beta$ be the angles of a non-isosceles triangle $ ABC$ at points $ A$ and $ B$, respectively. Let the bisectors of these angles intersect opposing sides of the triangle in $ D$ and $ E$, respectively. Prove that the acute angle between the lines $ DE$ and $ AB$ isn't greater than $ \frac{|\alpha\minus{}\beta|}3$.

In a parallelogram $ABCD$, the bisector of $\angle A$ intersects $BC$ at $M$ and the extension of $DC$ at $N$. Let $O$ be the circumcircle of the triangle $MCN$. Prove that $\angle OBC = \angle ODC$

In triangle $ ABC$, internal bisector of angle $ A$ intersects with $ BC$ at $ D$. Let $ E$ be a point on $ \left[CB\right.$ such that $ \left|DE\right|\equal{}\left|DB\right|\plus{}\left|BE\right|$. The circle through $ A$, $ D$, $ E$ intersects $ AB$ at $ F$, again. If $ \left|BE\right|\equal{}\left|AC\right|\equal{}7$, $ \left|AD\right|\equal{}2\sqrt{7}$ and $ \left|AB\right|\equal{}5$, then $ \left|BF\right|$ is $\textbf{(A)}\ \frac {7\sqrt {5} }{5} \qquad\textbf{(B)}\ \sqrt {7} \qquad\textbf{(C)}\ 2\sqrt {2} \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ \sqrt {10}$

In triangle $ABC$ we have $AB = 2, AC = 6$ and $\angle A = 120^o$ . The bisector of angle $A$ intersects the side BC at the point $D$. Determine the length of $AD$. The answer must be given as a fraction with integer numerator and denominator.

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 $ ABC$ be a triangle in which $ AB \equal{} AC$ and let $ I$ be its in-centre. Suppose $ BC \equal{} AB \plus{} AI$. Find $ \angle{BAC}$

On the bisector of the angle $ BAC $ of the triangle $ ABC $ we choose the points $ {{B} _ {1}}, \, \, {{C} _ {1}} $ for which $ B {{B} _ {1 }}\perp AB $, $ C {{C} _ {1}} \perp AC $. The point $ M $ is the midpoint of the segment $ {{B} _ {1}} {{C} _ {1}} $. Prove that $ MB = MC $.