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# In the given figure, if $∆ABE \cong ∆ACD$, show that $∆ADE \sim ∆ABC$.

"

Given:

$∆ABE \cong ∆ACD$

To do:

We have to show that $∆ADE \sim ∆ABC$.

Solution:

$\Delta ABE \cong \triangle ACD$

$AB=AC$ (CPCT)

$AE=AD$ (CPCT)

Therefore,

$\frac{AB}{AC}=\frac{AD}{AE}=\frac{1}{1}$

$\angle DAE=\angle BAC$

Therefore, by SAS criterion,

$\Delta ADE \sim \Delta ABC$

Hence proved.

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