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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