Let $ABC$ be an isosceles right-angled triangle of area $1$. Find the length of the shortest segment that divides the triangle into two parts of equal area.
Let $ABCD$ be a parallelogram and points $E,F$ be on its exterior. If triangles $BCF$ and $DEC$ are similar, i.e. $\triangle BCF\sim\triangle DEC$, prove that triangle $AEF$ is similar to these two triangles.
Seven triangles of area $7$ lie in a square of area $27$. Prove that among the $7$ triangles there are $2$ that intersect in a region of area not less than $1$.