Question

In the equivalent triangle ABC,D is a point on side BC such that BD=1/3.prove that 9AD^2=7AB^2.​

Answer 1

Answer:

Construction: Draw AP ⟂ BC

In Δ ADP,

AD2 = AP2 + DP2

AD2 = AP2 + (BP - BD)2

AD2 = AP2 + BP2 + BD2 - 2(BP)(BD)

AD2 = AB2 + (1/3 BC)2 - 2(BC/2)(BC/3)[AP^2+BP^2=AB^2(Pythagoras theorem)

AD2 = 7/9 AB2 (Since BC = AB)

9AD2 = 7AB^2