Q11 of 95 Page 6

In the given figure DB ⊥ BC, DE ⊥ AB and AC ⊥ BC, prove that ΔBDE ~ΔABC.


We have, DB BC and AC BC


∠B + ∠C = 90° + 90°


⇒ ∠B + ∠C = 180°


∴ BD || AC


⇒∠EBD = ∠CAB (alternate angles)


Let us take BDE andABC


∠BED = ∠ACB (each 90°)


∠EBD = ∠CAB (alternate angles)


∴ BDE ~ABC (by AA similarity criterion)


Hence Proved


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