In the given fig, ∆ABD is a right triangle, right angled at A and AC ⊥ BD. Prove that AB2 = BC. BD.
In the given fig, express x in terms of a, b and c.
In ∆KPN and ∆KLM, we have
∠KNP = ∠KML = 46° [Given]
∠K = ∠K [Common]
∴
[Using AA similar condition]
[∵ Corresponding sides of similar triangles are proportional]