M L Aggarwal Solutions for Chapter: Triangles, Exercise 8: 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.

Prove that the area of the semicircle drawn on the hypotenuse of a rightangled triangle is equal to the sumof the areas of the semicircles drawn on the other two sides of the triangle.

Prove that the area of the equilateral triangle drawn on the hypotenuse of a right-angled triangle is equal to the sum of the areas of the semicircles drawn on the other two sides of the triangle.