Question

the perpendicular from A on side BC of a Triangle ABC intersects BC at D such that DB=1/3CD. Prove that 2AB^2= 2AC^2 +BC2

Answer 1

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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.

Answer 2

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