In the figure, E is a point on side CB produced of an isosceles ∆ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ABD ~ ∆ECF
Given: A ∆ABC in which AB = AC and AD ⊥ BC. Side CB is produced to E and EF ⊥ AC.
To prove: ∆ABD ~ ∆ECF.
Proof: we known that the angles opposite to equal sides of a triangle are equal.
∠B = ∠C [∵ AB = AC]
Now, in ∆ABD and ∆ECF, we have
∴ ∠B = ∠C [proved above]
∠ADB = ∠EFC = 90º
∴ ∆ABD ~ ∆ECF [By AA-similarity]
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