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.