Question

In the adjoining figure, $D$ is a point on $BC$ such that $\angle ABD=\angle CAD$. If $AB=5cm,AC=3cm$ and $AD=4cm$, find
(i) $BC$
(ii) $DC$
(iii) area of $△ACD$: area of $△BCA$.

Answer 1

From the question it is given that,
$\angle ABD=\angle CAD$
$AB=5cm,AC=3cm$ and $AD=4cm$
Now, consider the $△ABC$ and $△ACD$
$\angle C=\angle C$ … [common angle for both triangles]
$\angle ABC=\angle CAD$ … [from the question]
So, $△ABC\sim △ACD$
Then, $AB/AD=BC/AC=AC/DC$
(i) Consider $AB/AD=BC/AC$
$5/4=BC/3$
$BC=(5\times 3)/4$
$BC=15/4$
$BC=3.75cm$
(ii) Consider $AB/AD=AC/DC$
$5/4=3/DC$
$DC=(3\times 4)/5$
$DC=12/5$
$Dc=2.4cm$
(iii) Consider the $△ABC$ and $△ACD$
$\angle CAD=\angle ABC$ … [from the question]
$\angle ACD=\angle ACB$ … [common angle for both triangle]
Therefore, $△ACD\sim △ABC$
Then, area of $△ACD$/area of $△ABC=A{D}^2/A{B}^2$
$=4^2/5^2$
$=16/25$
Therefore, area of $△ACD$ : area of $△BCA$ is $16:25$.