Question

an equilateral  ABC, D is a point on side BC such that BD = 13 BC. Prove that 9(AD)2 = 7(AB)2

Answer 1

Hope this will help you

Answer 2

Answer:

Given = BD = 1/3 BC

To prove = 9AD^2 = 7AB^2

CONSTRUCTION = We draw AE ⊥ BC

Proof :- Let the sides of the equilateral triangle be a and AE be the altitude of

Δ ABC

∴ BE = EC = BC/2 = a/2

And AE = a √3/2

ALSO,

BD = 1/3BC

∴ DE = BE-BD

⇒ a/2 -a/3 = a/6

BY APPLYING PYTHAGORAS THEOREM

AD^2 = AE^2+DC^2

AD^2 = (a √3/2)^2+ (a/6)

AD^2 = (3a^2/4) + (a/36)

AD^2 =  28a^2/36

AD^2 = 7AB^2/9

9AD^2 = 7AB^2

                                                                                                                               

HOPE IT HELPS YOU