Higher Order Thinkng Skills (HOTS)

In the adjoining figure, line segment $DF$ intersect the side $AC$ of triangle $ABC$ at the point $E$ such that $E$ is mid-point of $CA$ and $\angle AEF=\angle AFE$. Prove that $\frac{BD}{CD}=\frac{BF}{CE}$.

In the adjoining figure, $ABC$ is a triangle right-angled at $B$ and $BD\perp AC.$ If $AD=4 cm$ and $CD=5 cm$, find $BD$ and $AB.$

In the adjoining figure, $ABC$ is an isosceles triangle with $AB=AC=13 cm$ and $BC=10cm$. If $EFGH$ is a square inscribed in $△ABC$, find the area of the shaded region.