Question

In the trapezium $ABCD,AB\parallel DC$ and $△AED\sim △BEC$. Prove that $AD=BC.$

Answer 1

To Prove: $AD=BC$

Proof: Compare $△EDC$ and $△EBA$

$\angle EDC=\angle EBA$ [Alternate Angles]

$\angle ECD=\angle EAB$ [Alternate Angles]

By $AA$ criterion of similarity,
$△EDC\sim △EBA$

$\Rightarrow \frac{ED}{EB}=\frac{EC}{EA}$ [By CPST]

$\Rightarrow \frac{ED}{EC}=\frac{EB}{EA}\dots \left(i\right)$

But, $△AED\sim △BEC$ [Given]

$\Rightarrow \frac{ED}{EC}=\frac{EA}{EB}=\frac{AD}{BC}$ $\dots \left(ii\right)$ [By CPST]

$\frac{EB}{EA}=\frac{EA}{EB}$ [From $\left(i\right)$ and $\left(ii\right)$]

$\Rightarrow E{A}^2=E{B}^2$

$\Rightarrow EA=EB$

Hence, from equation $\left(ii\right),$
$\frac{AD}{BC}=1$

$\Rightarrow AD=BC$

Hence proved.