Let $G$ be the point of intersection of the medians in the triangle $ABC$. Let us denote $A_1, B_1, C_1$ the second points of intersection of lines $AG, BG, CG$ with the circle circumscribed around the triangle. Prove that $AG + BG + CG \le A_1C + B_1C + C_1C$. (Yasinsky V.A.)

Given a convex pentagon $ABCDE$ in which $\angle ABC = \angle AED = 90^o$, $\angle BAC= \angle DAE$. Let $K$ be the midpoint of the side $CD$, and $P$ the intersection point of lines $AD$ and $BK$, $Q$ be the intersection point of lines $AC$ and $EK$. Prove that $BQ = PE$.

The point $P$ is outside the circle $\omega$ with center $O$. Lines $\ell_1$ and $\ell_2$ pass through a point $P$, $\ell_1$ touches the circle $\omega$ at the point $A$ and $\ell_2$ intersects $\omega$ at the points $B$ and $C$. Tangent to the circle $\omega$ at points $B$ and $C$ intersect at point $Q$. Let $K$ be the point of intersection of the lines $BC$ and $AQ$. Prove that $(OK) \perp (PQ)$.

Consider the triangle $ABC$, in which $AB > AC$. Let $P$ and $Q$ be the feet of the perpendiculars dropped from the vertices $B$ and $C$ on the bisector of the angle $BAC$, respectively. On the line $BC$ note point $B$ such that $AD \perp AP.$ Prove that the lines $BQ, PC$ and $AD$ intersect at one point.

Two different circles $\omega_1$ ,$\omega_2$, with centers $O_1, O_2$ respectively intersect at the points $A, B$. The line $O_1B$ intersects $\omega_2$ at the point $F (F \ne B)$, and the line $O_2B$ intersects $\omega_1$ at the point $E (E\ne B)$. A line was drawn through the point $B$, parallel to the $EF$, which intersects $\omega_1$ at the point $M (M \ne B)$, and $\omega_2$ at the point $N (N\ne B)$. Prove that the lines $ME, AB$ and $NF$ intersect at one point.

Given a triangle $ABC$, the line passing through the vertex $A$ symmetric to the median $AM$ wrt the line containing the bisector of the angle $\angle BAC$ intersects the circle circumscribed around the triangle $ABC$ at points $A$ and $K$. Let $L$ be the midpoint of the segment $AK$. Prove that $\angle BLC=2\angle BAC$.

Let $ABC$ be an isosceles triangle with $AB = AC$. Let $D$ be a point on the base $BC$ such that $BD:DC = 2: 1$. Note on the segment $AD$ a point $P$ such that $\angle BAC= \angle BPD $. Prove that $\angle BPD = 2 \angle CPD$.

Given a triangle $ABC$. Point $M$ moves along the side $BA$ and point $N$ moves along the side $AC$ beyond point $C$ such that $BM=CN$. Find the geometric locus of the centers of the circles circumscribed around the triangle $AMN$.

The polyhedron $PABCDQ$ has the form shown in the figure. It is known that $ABCD$ is parallelogram, the planes of the triangles of the $PAC$ and $PBD$ mutually perpendicular, and also mutually perpendicular are the planes of triangles $QAC$ and $QBC$. Each face of this polyhedron is painted black or white so that the faces that have a common edge are painted in different colors. Prove that the sum of the squares of the areas of the black faces is equal to the sum of the squares of the areas of the white faces.

Given a right triangle $ABC$ with $ \angle C=90^o$. On its hypotenuse $AB$ is arbitrary mark the point$ P$. The point $Q$ is symmetric to the point $P$ wrt $AC$. Let the lines $PQ$ and $BQ$ intersect $AC$ at points $O$ and $R$ respectively. Denote by $S$ the foot of the perpendicular from the point $R$ on the line $AB$ ($S \ne P$), and let $T$ be the intersection point of lines $OS$ and $BR$. Prove that $R$ is the center of the circle inscribed in the triangle $CST$.

Given a quadrangular pyramid $SABCD$, the basis of which is a convex quadrilateral $ABCD$. It is known that the pyramid can be tangent to a sphere. Let $P$ be the point of contact of this sphere with the base $ABCD$. Prove that $\angle APB + \angle CPD = 180^o$.

On the sides $AB$ and $BC$ arbitrarily mark points $M$ and $N$, respectively. Let $P$ be the point of intersection of segments $AN$ and $BM$. In addition, we note the points $Q$ and $R$ such that quadrilaterals $MCNQ$ and $ACBR$ are parallelograms. Prove that the points $P,Q$ and $R$ lie on one line.

Let $ABC$ be an isosceles triangle in which $AB = AC$. On its sides $BC$ and $AC$ respectively are marked points $P$ and $Q$ so that $PQ\parallel AB$. Let $F$ be the center of the circle circumscribed about the triangle $PQC$, and $E$ the midpoint of the segment $BQ$. Prove that $\angle AEF = 90^o $.

The height $SO$ of a regular quadrangular pyramid $SABCD$ forms an angle $60^o$ with a side edge , the volume of this pyramid is equal to $18$ cm$^3$ . The vertex of the second regular quadrangular pyramid is at point $S$, the center of the base is at point $C$, and one of the vertices of the base lies on the line $SO$. Find the volume of the common part of these pyramids. (The common part of the pyramids is the set of all such points in space that lie inside or on the surface of both pyramids).

About the triangle $ABC$ it is known that $AM$ is its median, and $\angle AMC = \angle BAC$. On the ray $AM$ lies the point $K$ such that $\angle ACK = \angle BAC$. Prove that the centers of the circumcircles of the triangles $ABC, ABM$ and $KCM$ lie on the same line.

On the base of the $ABC$ of the triangular pyramid $SABC$ mark the point $M$ and through it were drawn lines parallel to the edges $SA, SB$ and $SC$, which intersect the side faces at the points $A1_, B_1$ and $C_1$, respectively. Prove that $\sqrt{MA_1}+ \sqrt{MB_1}+ \sqrt{MC_1}\le \sqrt{SA+SB+SC}$

Given a triangle $ABC$. Let $\Omega$ be the circumscribed circle of this triangle, and $\omega$ be the inscribed circle of this triangle. Let $\delta$ be a circle that touches the sides $AB$ and $AC$, and also touches the circle $\Omega$ internally at point $D$. The line $AD$ intersects the circle $\Omega$ at two points $P$ and $Q$ ($P$ lies between $A$ and $Q$). Let $O$ and $I$ be the centers of the circles $\Omega$ and $\omega$. Prove that $OD \parallel IQ$.

Let $t$ be a line passing through the vertex $A$ of the equilateral $ABC$, parallel to the side $BC$. On the side $AC$ arbitrarily mark the point $D$. Bisector of the angle $ABD$ intersects the line $t$at the point $E$. Prove that $BD=CD+AE$.

Let $AD$ be the bisector of triangle $ABC$. Circle $\omega$ passes through the vertex $A$ and touches the side $BC$ at point $D$. This circle intersects the sides $AC$ and $AB$ for the second time at points $M$ and $N$ respectively. Lines $BM$ and $CN$ intersect the circle for the second time $\omega$ at points $P$ and $Q$, respectively. Lines $AP$ and $AQ$ intersect side $BC$ at points $K$ and $L$, respectively. Prove that $KL=\frac12 BC$

The vertex $F$ of the parallelogram $ACEF$ lies on the side $BC$ of parallelogram $ABCD$. It is known that $AC = AD$ and $AE = 2CD$. Prove that $\angle CDE = \angle BEF$.

The quadrilateral $ABCD$ is inscribed in the circle and the lengths of the sides $BC$ and $DC$ are equal, and the length of the side $AB$ is equal to the length of the diagonal $AC$. Let the point $P$ be the midpoint of the arc $CD$, which does not contain point $A$, and $Q$ is the point of intersection of diagonals $AC$ and $BD$. Prove that the lines $PQ$ and $AB$ are perpendicular.