Question

5 marks question.....

Answer 1

$\textbf{Answer is in Attachment !}$

Answer 2

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 / 1616(AB2  - AC2 ) = 8BC2 2(AB2  - AC2 ) = BC2 2AB2 = 2AC2 + BC2 Hence proved.