Question

If $\Delta ABE\cong \Delta ACD$, show that $\Delta ADE\sim \Delta ABC$ .

Answer 1

Given $△ABE\cong △ACD$

As $△ABE=△ACD$

By corresponding points of conguent ${△}^{\prime }s$

$AB=AC$

$\Rightarrow \angle ABC=\angle ACB\Theta$

Also, $AE=AD$

$\Rightarrow \angle ADE=\angle AED=\alpha$

In $△ADE$ by angle sum property

$\angle ADE+\angle DEA+\angle DAE={180}^o$

$\angle DAE={100}^o-20---\left(1\right)$

In $△ABC$, by Angle sum property

$\angle BAC={180}^o-2\alpha$

$BAC={180}^o-2\alpha --\left(2\right)$

From equation $\left(1\right)$ & equation $\left(2\right)$

${180}^o-20={180}^o-2\alpha$

$\Rightarrow \alpha =0$

So, $\angle ADE=\angle ABC$

As corresponding Angle are equal

$\Rightarrow DE||BC$

So, by convense of Base proportionality theorem

$\frac{AD}{AB}=\frac{AE}{AC}$

In $△ADE$ and $△ABC$

$\frac{AD}{AB},\frac{AE}{AC}$ and $\angle DAE=\angle BAC$

so $△ADE\sim △ABC$ [by $SAS$ similarity rule]