Školjka

%V0 Given a triangle $ABC$, let $D$ and $E$ be points on the side $BC$ such that $\angle BAD = \angle CAE$. If $M$ and $N$ are, respectively, the points of tangency of the incircles of the triangles $ABD$ and $ACE$ with the line $BC$, then show that $$\frac{1}{MB}+\frac{1}{MD}= \frac{1}{NC}+\frac{1}{NE}.$$

%V0 The incircle of triangle $\triangle ABC$ touches the sides $BC$, $CA$, $AB$ at $D, E, F$ respectively. $X$ is a point inside triangle of $\triangle ABC$ such that the incircle of triangle $\triangle XBC$ touches $BC$ at $D$, and touches $CX$ and $XB$ at $Y$ and $Z$ respectively. Show that $E, F, Z, Y$ are concyclic.

%V0 An acute triangle $ABC$ is given. Points $A_1$ and $A_2$ are taken on the side $BC$ (with $A_2$ between $A_1$ and $C$), $B_1$ and $B_2$ on the side $AC$ (with $B_2$ between $B_1$ and $A$), and $C_1$ and $C_2$ on the side $AB$ (with $C_2$ between $C_1$ and $B$) so that $$\angle AA_1A_2 = \angle AA_2A_1 = \angle BB_1B_2 = \angle BB_2B_1 = \angle CC_1C_2 = \angle CC_2C_1.$$ The lines $AA_1,BB_1,$ and $CC_1$ bound a triangle, and the lines $AA_2,BB_2,$ and $CC_2$ bound a second triangle. Prove that all six vertices of these two triangles lie on a single circle.

%V0 Circles $S_1$ and $S_2$ intersect at points $P$ and $Q$. Distinct points $A_1$ and $B_1$ (not at $P$ or $Q$) are selected on $S_1$. The lines $A_1P$ and $B_1P$ meet $S_2$ again at $A_2$ and $B_2$ respectively, and the lines $A_1B_1$ and $A_2B_2$ meet at $C$. Prove that, as $A_1$ and $B_1$ vary, the circumcentres of triangles $A_1A_2C$ all lie on one fixed circle.

%V0 In an acute triangle $ABC$ segments $BE$ and $CF$ are altitudes. Two circles passing through the point $A$ anf $F$ and tangent to the line $BC$ at the points $P$ and $Q$ so that $B$ lies between $C$ and $Q$. Prove that lines $PE$ and $QF$ intersect on the circumcircle of triangle $AEF$. Proposed by Davood Vakili, Iran

%V0 Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram. Prove that $GR=GS$. Proposed by Hossein Karke Abadi, Iran