Question

In ∆ABC, ∠B = 90° and BD ⊥ AC. Prove that ∠ABD = ∠ACB.

Answer 1

Side AD of triangle ABD is produced to C.
$\therefore \angle BDC=\angle BAD+\angle ABD...\left(i\right)$
From the right $∆ABC$, we have:
$$\begin{gathered} \angle ABC=\angle BAC+\angle ACB \\ \Rightarrow 90°=\angle BAD+\angle ACB \\ \Rightarrow 90°=\angle BDC-\angle ABD+\angle ACB\left[\text{Using}\left(i\right)\right] \\ \Rightarrow 90°=90°-\angle ABD+\angle ACB\left[\because BD\perp AC\right] \\ \Rightarrow \angle ABD=\angle ACB \end{gathered}$$