Found problems: 10
In the right triangle $ABC$ with hypotenuse $AB$, the incircle touches $BC$ and $AC$ at points ${{A}_{1}}$ and ${{B}_{1}}$ respectively. The straight line containing the midline of $\Delta ABC$ , parallel to $AB$, intersects its circumcircle at points $P$ and $T$. Prove that points $P,T,{{A}_{1}}$ and ${{B}_{1}}$ lie on one circle.
In a triangle $ABC$, let $D$ and $E$ be points of the sides $ BC$ and $AC$ respectively. Segments $AD$ and $BE$ intersect at $O$. Suppose that the line connecting midpoints of the triangle and parallel to $AB$, bisects the segment $DE$. Prove that the triangle $ABO$ and the quadrilateral $ODCE$ have equal areas.
In a triangle $ABC$, let $D$ and $E$ be points of the sides $ BC$ and $AC$ respectively. Segments $AD$ and $BE$ intersect at $O$. Suppose that the line connecting midpoints of the triangle and parallel to $AB$, bisects the segment $DE$. Prove that the triangle $ABO$ and the quadrilateral $ODCE$ have equal areas.
Let $ABC$ be a triangle. It is known that the triangle formed by the midpoints of the medians of $ABC$ is equilateral. Prove that $ABC$ is equilateral as well.
In the acute triangle $ABC$, the altitude $AH$ is drawn. Using segments $AB,BH,CH$ and $AC$ as diameters circles $\omega_1, \omega_2, \omega_3$ and $\omega_4$ are constructed respectively. Besides the point $H$, the circles $\omega_1$ and $\omega_3$ intersect at the point $P,$ and the circles $\omega_2$ and $\omega_4$ interext at point $Q$. The lines $BQ$ and $CP$ intersect at point $N$. Prove that this point lies on the midline of triangle $ABC$, which is parallel to $BC$.
A trapezoid is given in which one base is twice as large as the other. Use one ruler (no divisions) to draw the midline of this trapezoid.
In an acute non-isosceles triangle $ABC$, the altitude $AA'$ is drawn and point $H$ is the intersection point of the altitudes and and $O$ is the center of the circumscribed circle. Prove that the point symmetric to the circumcenter of triangle $HOA'$ wrt straight line $HO$, lies on a midline of triangle $ABC$.
In a convex hexagon $ABCDEF$ the triangles $BDF, ACE$ are equilateral and congruent. Prove that the three lines connecting the midpoints of opposite sides are concurrent.
The line $\ell$ is perpendicular to the side $AC$ of the acute triangle $ABC$ and intersects this side at point $K$, and the circumcribed circle $\vartriangle ABC$ at points $P$ and $T$ (point P on the other side of line $AC$, as the vertex $B$). Denote by $P_1$ and $T_1$ - the projections of the points $P$ and $T$ on line $AB$, with the vertices $A, B$ belong to the segment $P_1T_1$. Prove that the center of the circumscribed circle of the $\vartriangle P_1KT_1$ lies on a line containing the midline $\vartriangle ABC$, which is parallel to the side $AC$.
(Anton Trygub)
Points $A_1$, $B_1$, $C_1$ are midpoints of sides $BC$, $AC$, $AB$ of triangle $ABC$. On midlines $C_1B_1$ and $A_1B_1$ points $E$ and $F$ are chosen such that $BE$ is the angle bisector of $AEB_1$ and $BF$ is the angle bisector of $CFB_1$. Prove that bisectors of angles $ABC$ and $FBE$ coincide.
[I]Proposed by F. Baharev[/i]