Question

In figure, if $\Delta$ABE $\cong \Delta$ACD, show that $\Delta$ADE $-\Delta$ABC.

Answer 1

REF.Image

Given $\Delta ABE\cong \Delta ACD$

To prove $\Delta ADE\sim \Delta ABC$

proof it is given that

$\Delta ABE\cong \Delta ACD$

Therefore AB= AC [By CPCT] _______ (1)

And AE=AD [By CPCT]

$\Rightarrow AD=AE$ __________ (2)

Now Divide (2) by (1) we have

$\frac{AD}{AB}=\frac{AE}{AC}$ _________ (3)

Now in $\Delta ADE$ & $\Delta ABC$

$\frac{AD}{AB}=\frac{AE}{AC}$ [ from equation (3)]

$\angle A=\angle A$ [ common]

$\Rightarrow \Delta ADE\sim \Delta ABC$

By SAS criterian of similarity !

Hence proved !