Question

________________________
Answer this properly :-
________________________
===========================
The perpendicular from A on side BC of a triangle ABC intersects BC at D such that DB= 3CD .prove that 2AB²= 2AC²+BC²
===========================

​

Answer 1

Step-by-step explanation:

SOL......

Given that AD ⊥ BC and DB = 3CD To prove : 2AB2 = 2AC2 + BC2

Proof : BD + DC = BC 3CD + CD = BC 4CD = BC ⇒ CD = BC / 4 DB = 3CD = 3BC / 4. In a right angle traingle ACD , AC2 = AD2 + CD2. AC2 = AD2 + BC2 / 16 -------(1) In a right angle traingle ABD , AB2 = AD2 + BD2. AB2 = AD2 + 9BC2 / 16 -------(2). Substracting (1) from (2) we obtain AB2 - AC2 = 9BC2 / 16 - BC2 / 16 16(AB2 - AC2 ) = 8BC2 2(AB2 - AC2 ) = BC2 2AB2 = 2AC2 + BC2 Hence proved.