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In the figure, if \(\triangle\) ABE \({\displaystyle \cong }\) \(\triangle\) ACD, show that \(\triangle\) ADE ~ \(\triangle\) ABC.
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Answer :
Given, \(\triangle\) ABE \({\displaystyle \cong }\) \(\triangle\) ACD.
AB = AC [By CPCT] ……………………………….(i)
And, AD = AE [By CPCT] ……………………………(ii)
In \(\triangle\) ADE and \(\triangle\) ABC, dividing eq.(ii) by eq(i),
AD/AB = AE/AC
\(\angle\) A = \(\angle\) A [Common angle]
\(\triangle\) ADE ~ \(\triangle\) ABC [SAS similarity criterion]