Question

In the adjoining figure, $D$ is a point on $BC$ such that $\angle ABD=\angle CAD$. If $AB=5\text{cm},AD=4\text{cm}$ and $AC=3\text{cm}$. Find $A\left(△ACD\right):A\left(△BCA\right)$

Answer 1

In $△ABC$ and $△CAD$,
$\angle ABD=\angle CAD$ (Given)
$\angle ACB=\angle ACD$ (Common angle)
$\angle CAB=\angle CDA$ (Third angle - By AA Test of Similarity)
Hence,

$△ABC\sim △DAC$ (AAA rule)

By Theorem of Areas of Similar Triangles, ratios of areas of similar triangles is equal to the ratio of square of corresponding sides
$\frac{Ar.△ACD}{Ar.△BCA}=\frac{A{D}^2}{A{B}^2}$
$\frac{Ar.△ACD}{Ar.△BCA}=\frac{4^2}{5^2}$
$\frac{Ar.△ACD}{Ar.△BCA}=\frac{16}{25}$