Question

In an equalata triangle D is a point on side BC such that 4BD=BC. Prove that 16AD square = 13 BC square

Answer 1

Draw AE, where E is the midpoint of BC. AE is perpendicular to BC.

BD = BC/4. BE=BC/2. DE= BC/4
AD^2 = DE^2 + AE^2
= BC^2/4^2 + 3 BC^2/4
= 13 BC^2 /16

Hence the answer.