Q10 of 32 Page 6

In the given figure, DB ⊥ BC, DE ⊥ AB, and AC ⊥ BC.

Prove that


(CBSE 2008)

Observe in ∆BED & ∆ACB, we have

∠BED = ∠ACB = 90°


Now according to what’s given, DB ┴ BC and AC ┴ BC we can write,


∠B + ∠C = 180°


This clearly means BD ∥ CA


⇒ ∠EBD = ∠CAB [They are alternate angles]


Thus, by AA-similarity theorem, ∆BED ∼ ∆ACB


So,


⇒


Hence, proved.

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