Question

In $△ABC,\angle ABC=\angle DAC,AB=8cm,AC=4cm,AD=5cm$.

Find $\frac{\text{area of}△ABC}{\text{area of}△ACD}$

Answer 1

In $△ACD$ and $△BCA$
$\angle C=\angle C$ (common)
$\angle ABC=\angle CAD$ (given)
$\therefore △ACD\sim △BCA$ ($A-A$ postulates)

$\therefore \frac{\text{area}\left(△ACD\right)}{\text{area}\left(△ABC\right)}=\frac{A{C}^2}{A{B}^2}$
$=\frac{(4{)}^2}{(8{)}^2}=\frac{16}{64}=\frac{1}{4}$
$\text{area of}△ABC:\text{area of}△ACD=4:1$