Question

. In D ABC, seg AD ^ seg BC

DB = 3CD. Prove that ः

2AB2

= 2AC2

+ BC2​

Answer 1

Hey mate your answer is here ⬇️⬇️

Given that in ΔABC, we have

AD ⊥BC and BD = 3CD

In right angle triangles ADB and ADC, we have

AB2 = AD2 + BD2 ...(i)

AC2 = AD2 + DC2 ...(ii) [By Pythagoras theorem]

Subtracting equation (ii) from equation (i), we get

AB2 - AC2 = BD2 - DC2

= 9CD2 - CD2 [∴ BD = 3CD]

= 9CD2 = 8(BC/4)2 [Since, BC = DB + CD = 3CD + CD = 4CD]

Therefore, AB2 - AC2 = BC2/2

⇒ 2(AB2 - AC2) = BC2

⇒ 2AB2 - 2AC2 = BC2

∴ 2AB2 = 2AC2 + BC2.