Question

In Δ ABC (Fig. 11.23), AD is ⊥ BC, BE is ⊥ AC and AD = BE. Prove that AE = BD.

Answer 1

area of ∆ADB = ∆ BEC -------- (1) | R.H.S theorem

In ∆ ADB & ∆ AEB

angADB = angAEB | 90°

∆ADB =(congrent) ∆ AEB

AD = AE ----- (2) { c. p. c. t. }

from (1) & (2)

AE = BE

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