Q15 of 186 Page 211

In the given figure, ∠1 = ∠2 and . Prove that ΔACB ~ ΔDCE.

To Prove: ∆ACB ∼ ∆DCE

Proof:


Given that, ∠1 = ∠2


⇒ ∠DBC = ∠DCE


Also in ∆ABC & ∆DCE, we get


∠DCE = ∠ACB [they are common angles to both triangles]


And


Or


Or [∵ BD = DC as ∠1 = ∠2]


Thus by SAS-similarity criteria, we get


∆ACB ∼ ∆DCE


Hence, proved.


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