Question

ΔABC is an equilateral triangle. Point D is on seg BC such that BD = `1/5` BC. Then prove that `(AD^2)/(AB^2) = 21/25`.

Answer 1

Construction: Draw AP ⊥ side BC.

Proof: △ABC is an equilateral triangle.

∴ ∠B = 60° and AB = AC = BC.

∠APB = 90°       ...(Construction)

∴ ∠BAP = 30°     ...(The remaining angle of △ABP.)

∴ △ABP is a 30° − 60° − 90° triangle. 

∴ AP = `(sqrt3)/2` AB      ...(The side opposite to 60°)    ...(1)

and BP = `1/2 "AB" = 1/2 "BC"`     ...(The side opposite to 30°)   ...(2)

DP = BP − BD        ...(B − D − P)

= `1/2 "BC" - 1/5 "BC"`      ...[From (2) and given]

= `(5 "BC" - 2"BC")/10 = 3/10 "BC"`.

= In △ADP, by Pythagorean theorem,

AD² = AP² + DP²

= `((sqrt3)/2 "AB")^2 + (3/10 "BC")^2`      ...[From (1) and (3)]

= `3/4 "AB" + 9/100 "AB"^2`     ...(∵ BC = AB)

= `(75 "AB"^2 + 9"AB"^2)/100`

∴ `"AD"^2 = (84"AB"^2)/100`

∴ `"AD"^2 = (21"AB"^2)/25`

∴ `("AD"^2)/("AB"^2) = 21/25`.