In triangle ABC, AD is perpendicular to side BC and AD² = BD × CD.

prove that angle BAC = 90°.

In the given figure, Δ ABC is isosceles and AP x BQ = AC², prove that Δ ACP ∼ Δ BCQ.

In triangle ABC, ∠BAC = 90° and AD is perpendicular to side BC. Triangle ABD is similar to triangle CBA by :

1. SAS

2. ASA

3. AAA

4. RHS