In an equilateral triangle ABC,D is a point on side BC such that BD=1(bc)/3, prove that 9AD^2=7AB^2

In an equilateral triangle ABC,D is a point on side BC such that BD=1(

1.	In an equilateral triangle ABC,D is a point on side BC such that BD=1(bc)/3, prove that 9AD^2=7AB^2
Answer» Given that ∆ABC is a equilateral ∆ so AB=CB=BA To prove = 9AD^2=7AB^2Proof-BD=bc/3 , dc=bc/2And , angle adc=adb=90°By pgt , Ab^2=ad^2+bd^2 , ad^2=Ab^2-bd^2 (1).. Similiarly, ad^2=ac^2-dc^2 (2)...From(1)&(2)..Ab^2-bd^2=ac^2-dc^2And, ab^2-(bc/3)^2=ac^2-(bc/2)^2After solving it we get , 9Ad^2=7ab^2