Question

In $\Delta ABC,\frac{AD}{DB}=\frac{AE}{EC}$ and $\angle ADE=\angle ACB.$

Then $\Delta ABC$ is a/an .

[ 3 Marks ]

Answer 1

By the Converse of Basic Proportionality theorem,
Since $\frac{AD}{DB}=\frac{AE}{EC}⟹DE\parallel BC$

(1 Mark)

Given $\angle ADE=\angle ACB...\left(1\right)$
Also, $\angle ADE=\angle ABC...\left(2\right)$
(Transverse angles in parallel lines are equal)

(1 Mark)

From (1) & (2),
$\Rightarrow \angle ABC=\angle ACB$
Hence it is an "isosceles triangle".

(1 Mark)