In the given figure, Δ ABC is isosceles and AP x BQ = AC2, prove that Δ ACP ∼ Δ BCQ.
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In triangle ABC, AD is perpendicular to side BC and AD2 = BD × CD.
prove that angle BAC = 90°.
In triangle ABC, ∠BAC = 90° and AD is perpendicular to side BC. Triangle ABD is similar to triangle CBA by :
SAS
ASA
AAA
RHS
If AE = 10 cm, BD = 8 cm and BC = 10 cm, then AB is equal to :
5 cm
25 cm
12.5 cm
2.5 cm