In the rectangle $ABCD$, $AB = 2BC$. An equilateral triangle $ABE$ is constructed on the side $AB$ of the rectangle so that its sides $AE$ and $BE$ intersect the segment $CD$. Point $M$ is the midpoint of $BE$. Find the $\angle MCD$.

It is known that the angles of the triangle $ABC$ are $1: 3: 5$. Find the angle between the bisector of the largest angle of the triangle and the line containing the altitude drawn to the smallest side of the triangle.

There is a ruler and a "rusty" compass, with which you can construct a circle of radius $R$. The point $K$ is from the line $\ell$ at a distance greater than $R$. How to use this ruler and this compass to draw a line passing through the point $K$ and perpendicular to line $\ell$? (Misha Sidorenko, Katya Sidorenko, Rodion Osokin)

The median $AM$ is drawn in the triangle $ABC$ ($AB \ne AC$). The point $P$ is the foot of the perpendicular drawn on the segment $AM$ from the point $B$. On the segment $AM$ we chose such a point $Q$ that $AQ = 2PM$. Prove that $\angle CQM = \angle BAM$.

It is known that a circle can be inscribed in the quadrilateral $ABCD$, in addition $\angle A = \angle C$. Prove that $AB = BC$, $CD = DA$. (Olena Artemchuk)

Let $ABCD$ be a square, point $E$ be the midpoint of the side $BC$. The point $F$ belongs to the side $AB$, and $DE \perp EF$. The point $G$ lies inside the square, and $GF = FE$ and $GF \perp FE$. Prove that: a) $DE$ is the bisector of the $\angle FDC$ b) $FG$ is the bisector of the $\angle AFD$ c) the point $G$ is the center of the circle inscribed in the triangle $ADF$. (Ercole Suppa, Italy)

Given a right triangle $ABC$, the point $M$ is the midpoint of the hypotenuse $AB$. A circle is circumscribed around the triangle $BCM$, which intersects the segment $AC$ at a point $Q$ other than $C$. It turned out that the segment $QA$ is twice as large as the side $BC$. Find the acute angles of triangle $ABC$. (Mykola Moroz)

An equilateral triangle $BDE$ is constructed on the diagonal $BD$ of the square $ABCD$, and the point $C$ is located inside the triangle $BDE$. Let $M$ be the midpoint of $BE$. Find the angle between the lines $MC$ and $DE$. (Dmitry Shvetsov)

Point $M$ is the midpoint of the side $CD$ of the trapezoid $ABCD$, point $K$ is the foot of the perpendicular drawn from point $M$ to the side $AB$. Give that $3BK \le AK$. Prove that $BC + AD\ge 2BM$.

Let $BB_1$ and $CC_1$ be the altitudes of the acute-angled triangle $ABC$. From the point $B_1$ the perpendiculars $B_1E$ and $B_1F$ are drawn on the sides $AB$ and $BC$ of the triangle, respectively, and from the point $C_1$ the perpendiculars $C_1 K$ and $C_1L$ on the sides $AC$ and $BC$, respectively. It turned out that the lines $EF$ and $KL$ are perpendicular. Find the measure of the angle $A$ of the triangle $ABC$. (Alexander Dunyak)

Let $AL$ be the bisector of triangle $ABC$. Circle $\omega_1$ is circumscribed around triangle $ABL$. Tangent to $\omega_1$ at point $B$ intersects the extension of $AL$ at point $K$. The circle $\omega_2$ circumscribed around the triangle $CKL$ intersects $\omega_1$ a second time at point $Q$, with $Q$ lying on the side $AC$. Find the value of the angle $ABC$. (Vladislav Radomsky)

In the triangle $ABC$ the altitude $BD$ and $CT$ are drawn, they intersect at the point $H$. The point $Q$ is the foot of the perpendicular drawn from the point $H$ on the bisector of the angle $A$. Prove that the bisector of the external angle $A$ of the triangle $ABC$, the bisector of the angle $BHC$ and the line $QM$, where $M$ is the midpoint of the segment $DT$, intersect at one point. (Matvsh Kursky)

The square $ABCD$ is divided into $8$ equal right triangles and the square $KLMN$, as shown in the figure. Find the area of the square $ABCD$ if $KL = 5, PS = 8$.

Let $ABCD$ be a square, point $E$ be the midpoint of the side $BC$. On the side $AB$ mark a point $F$ such that $FE \perp DE$. Prove that $AF + BE = DF$. (Ercole Suppa, Italy)

A trapezoid $ABCD$ with bases $BC$ and $AD$ is given. The points $K$ and $L$ are chosen on the sides $AB$ and $CD$, respectively, so that $KL \parallel AD$. It turned out that the areas of the quadrilaterals $AKLD$ and $KBCL$ are equal. Find the length $KL$ if $BC = 3, AD = 5$.

In an isosceles trapezoid $ABCD$, the base $AB$ is twice as large as the base $CD$. Point $M$ is the midpoint of $AB$. It is known that the center of the circle inscribed in the triangle $MCB$ lies on the circle circumscribed around the triangle $MDC$. Find the angle $\angle MBC$.

A cube whose edge is $1$ is intersected by a plane that does not pass through any of its vertices, and its edges intersect only at points that are the midpoints of these edges. Find the area of the formed section. Consider all possible cases. (Alexander Shkolny)

Given an acute triangle $ABC$. A circle inscribed in a triangle $ABC$ with center at point $I$ touches the sides $AB, BC$ at points $C_1$ and $A_1$, respectively. The lines $A_1C_1$ and $AC$ intersect at the point $Q$. Prove that the circles circumscribed around the triangles $AIC$ and $A_1CQ$ are tangent. (Dmitry Shvetsov)

On the midline $MN$ of the trapezoid $ABCD$ ($AD\parallel BC$) the points $F$ and $G$ are chosen so that $\angle ABF =\angle CBG$. Prove that then $\angle BAF = \angle DAG$. (Dmitry Prokopenko)

The segments $BF$ and $CN$ are the altitudes in the acute-angled triangle $ABC$. The line $OI$, which connects the centers of the circumscribed and inscribed circles of triangle $ABC$, is parallel to the line $FN$. Find the length of the altitude $AK$ in the triangle $ABC$ if the radii of its circumscribed and inscribed circles are $R$ and $r$, respectively. (Grigory Filippovsky)

The altitudes of the acute-angled triangle $ABC$ intersect at the point $H$. On the segments $BH$ and $CH$, the points $B_1$ and $C_1$ are marked, respectively, so that $B_1C_1 \parallel BC$. It turned out that the center of the circle $\omega$ circumscribed around the triangle $B_1HC_1$ lies on the line $BC$. Prove that the circle $\Gamma$, which is circumscribed around the triangle $ABC$, is tangent to the circle $\omega$ .

It is known about the triangle $ABC$ that $3 BC = CA + AB$. Let the $A$-symmedian of triangle $ABC$ intersect the circumcircle of triangle $ABC$ at point $D$. Prove that $\frac{1}{BD}+ \frac{1}{CD}= \frac{6}{AD}$. (Ercole Suppa, Italy)

In an isosceles triangle $ABC, I$ is the center of the inscribed circle, $M_1$ is the midpoint of the side $BC, K_2, K_3$ are the points of contact of the inscribed circle of the triangle with segments $AC$ and $AB$, respectively. The point $P$ lies on the circumcircle of the triangle $BCI$, and the angle $M_1PI$ is right. Prove that the lines $BC, PI, K_2K_3$ intersect at one point. (Mikhail Plotnikov)