Question

If D is a point on the side AB of $△$ ABC such that AD : DB = 3:2 and E is a point on BC such that DE $\parallel$ AC. Find the ratio of areas of $△$ ABC and $△$ BDE.

Answer 1

$\frac{AD}{DB}=\frac{3}{2}$ [ Given ]

$\therefore$ $\frac{AD}{DB}+1=\frac{3}{2}+1$

$\therefore$ $\frac{AD+DB}{DB}=\frac{3+2}{2}$

$\therefore$ $\frac{AB}{DB}=\frac{5}{2}$ --- ( 1 )

In $△BDE$ and $△ABC$

$\angle B=\angle B$ [ Common ]

$\angle BAC=\angle BDE$ [ $\because DE\parallel AC$ ]

$\therefore$ $△ABC\sim △BDE$ [ By AA similarity criteria ]

$\frac{ar\left(△ABC\right)}{ar\left(△BDE\right)}=\frac{A{B}^2}{D{B}^2}=\frac{5^2}{2^2}=\frac{25}{4}$ [ From ( 1 ) ]

$\therefore$ $ar\left(△ABC\right):ar\left(△BDE\right)=25:4$