Question

16. In an equilateral triangle ABC, D is a point on the side BC such that BD = BC. prove that 9 AD = 1 AB​

Answer 1

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)

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

9AD2 = 7AB2