44
In the given figure $\angle ADE = \angle ACB$ and $\frac{AD}{DB} = \frac{AE}{EC}$. Prove that $\Delta ABC$ is an isosceles triangle.
Show SolutionHide Solution↓
Solution: Since $\frac{AD}{DB} = \frac{AE}{EC}$ therefore $DE \parallel BC$ (By converse of BPT) [1/2 mark]
$\Rightarrow \angle ADE = \angle ABC$ (corresponding angles) [1/2 mark]
given $\angle ADE = \angle ACB$ therefore $\angle ABC = \angle ACB$ [1/2 mark]
$\Rightarrow AC = AB$ or $\Delta ABC$ is an isosceles triangle. [1/2 mark]
$\Rightarrow \angle ADE = \angle ABC$ (corresponding angles) [1/2 mark]
given $\angle ADE = \angle ACB$ therefore $\angle ABC = \angle ACB$ [1/2 mark]
$\Rightarrow AC = AB$ or $\Delta ABC$ is an isosceles triangle. [1/2 mark]