Problems

We call a median the segment from the vertex of a triangle to the midpoint of the opposite side. Prove that in two congruent triangles, the corresponding medians are of equal length.

We call a bisector the segment from the vertex of a triangle to the opposite side which divides in half the angle next to the starting vertex. Prove that in two congruent triangles, the corresponding bisectors are of equal length.

$7$ identical hexagons are arranged in a pattern on the picture below. Each hexagon has an area of $8$. What is the area of the triangle $\mathrm{△}ABC$?

In the triangle $ABC$ the bisector $BD$ coincides with the height. Prove that $AB=BC$.

In the triangle $ABC$ the median $BD$ coincides with the height. Prove that $AB=BC$.

In the triangle $ABC$ with $BC=12$, the median $AE$ is perpendicular to the bisector $BD$. Find the length of $AB$.

On the sides of the equilateral triangle $ABC$ three points $D,E,F$ are chosen in such a way that the following ratios of lengths hold: $$AD:DC=CF:FB=BE:AE$$ Prove that the triangle $DEF$ is also equilateral.

Two circles with centres $A$ and $C$ intersect at the points $B$ and $D$. Prove that the segment $AC$ is perpendicular to $BD$. Moreover, prove that the segment $AC$ divides $BD$ in half.

For two congruent triangles Prove that their corresponding heights are equal.

The sides $AB$ and $CD$ of the quadrilateral $ABCD$ are equal, the points $E$ and $F$ are the midpoints of $AB$ and $CD$ correspondingly. Prove that the perpendicular bisectors of the segments $BC$, $AD$, and $EF$ intersect at one point.