In trapezoid $ABCD$ with bases $AD, BC$, $AD = 2BC$ and $M$ is midpoint of $AB$. Prove that line $BD$ passes through the midpoint of segment $CM$.

There is a square sheet of paper. How to get a rectangular sheet of paper with an aspect ratio equal to $\sqrt2$? (There are no tools, the sheet can only be bent.)

Given is a triangle $ABC$ and $M$ is the midpoint of the minor arc $BC$. Let $M_1$ be the reflection of $M$ with respect to side $BC$. Prove that the nine-point circle bisects $AM_1$.

Let $I$ be the incenter of triangle $ABC$, tangent to sides $AB$ and $AC$ at points $E$ and $F$, respectively. The lines through $E$ and $F$ parallel to $AI$ intersect lines $BI$ and $CI$ at points $P$ and $Q$, respectively. Prove that the center of the circumcircle of triangle $IPQ$ lies on line $BC$.

Altitudes $BB_1$ and $CC_1$ of acute triangle $ABC$ intersect at $H$, and $\angle A = 60^{o}$, $AB < AC$. The median $AM$ intersects the circumcircle of $ABC$ at point $K$; $L$ is the midpoint of the arc $BC$ of the circumcircle that does not contain point $A$; lines $B_1C_1$ and $BC$ intersect at point $E$. Prove that $\angle EHL = \angle ABK$.

Given a circle $\Omega$ tangent to side $AB$ of angle $\angle BAC$ and lying outside this angle. We consider circles $w$ inscribed in angle $BAC$. The internal tangent of $\Omega$ and $w$, different from $AB$, touches $w$ at a point $K$. Let $L$ be the point of tangency of $w$ and $AC$. Prove that all such lines $KL$ pass through a fixed point without depending on the choice of circle $w$.

In triangle ABC $\angle ABC=60^{o}$ and $O$ is the center of the circumscribed circle. The bisector $BL$ intersects the circumscribed circle at the point $W$. Prove that $OW$ is tangent to $(BOL)$

Points $X_1$ and $X_2$ move along fixed circles with centers $O_1$ and $O_2$, respectively, so that $O_1X_1 \parallel O_2X_2$. Find the locus of the intersection point of lines $O_1X_2$ and $O_2X_1$.

In an acute triangle $ABC$ the line $OI$ is parallel to side $BC$. Prove that the center of the nine-point circle of triangle $ABC$ lies on the line $MI$, where $M$ is the midpoint of $BC$.

Given isosceles tetrahedron $PABC$ (faces are equal triangles). Let $A_0$, $B_0$ and $C_0$ be the touchpoints of the circle inscribed in the triangle $ABC$ with sides $BC$, $AC$ and $AB$ respectively, $A_1$, $B_1$ and $C_1$ are the touchpoints of the excircles of triangles $PCA$, $PAB$ and $PBC$ with extensions of sides $PA$, $PB$ and $PC$, respectively (beyond points $A$, $B$, $C$). Prove that the lines $A_0A_1$, $B_0B_1$ and $C_0C_1$ intersect at one point.

In an acute-angled triangle $ABC$ with orthocenter $H$, the line $AH$ cuts $BC$ at point $A_1$. Let $\Gamma$ be a circle centered on side $AB$ tangent to $AA_1$ at point $H$. Prove that $\Gamma$ is tangent to the circumscribed circle of triangle $AMA_1$, where $M$ is the midpoint of $AC$.

Points $C_1$ and $C_2$ lie on side $AB$ of triangle $ABC$, where the point $C_1$ belongs to the segment $AC_2$ and $\angle ACC_1= \angle BCC_2$. On segments $CC_1$ and $CC_2$ points $A'$ and $B'$ are taken such that $\angle CAA'= \angle CBB' = \angle C_1CC_2$. Prove that the center of the circle $(CA'B')$ lies on the perpendicular bisector of the segment $AB$.