Question

In triangle ABC,angleB=90°,D and E are the point on AC ,A-D-E-C,AD=DE=EC,prove that BDsquare+BEsquare=3DEsquare

Answer 1

Answer:

In Δ ABC, AB² + BC² = AC²

Given ∠B = 90°

Therefore, ∠A + ∠B + ∠C = 180°

∠B + ∠C = 90°

∠B = ∠C = 45°

Therefore, AB = AC

Now in Δ ABD and Δ DBE,

∠ abd = ∠ dbe = 30°

∠ADB = ∠ DEB = 105°

AD = DE

Therefore by  AAS similarity criterion, ΔABD is similar to Δ DBE

Therefore,  AB is similar to BD ........................ (1)

Similarly, Δ BEC is similar to ΔBDE

Therefore, BC is similar to BE ........................... (2)

Now use (1) and (2) in AB² + BC² = AC²

BD² + BE² = AC²

or, BD² + BE² = 3 DE² ( given that AD = DE = EC)

Hence proved.