ABC is an equilateral triangle D is a point on BC such that BD=1/3BC then prove that 9AB^2 = 7AD^2

ABC is an equilateral triangle D is a point on BC such that BD=1/3BC t

1.	ABC is an equilateral triangle D is a point on BC such that BD=1/3BC then prove that 9AB^2 = 7AD^2
Answer» Sorry i can not draw figure here\xa0Just drop perpendicular AE from A to BC\xa0suppose side of triangle is 6x thenAD=2x and DE=x and AB = 3x and BE=3xso AE2= AB2 -BE2= (6x)2 -(3x)2 =36x2-9x2 =27x2Now in triangle ADEAD2=AE2 + DE2 =27 x2\xa0+ x2\xa0 = 28 x2AD2\xa0= 28/36 36x2 = 28/36 AB2 = 7/9*AB27 AD2\xa0= 9 AB2\xa0\xa0\xa0\xa0 Sorry , show that 9AD^2 = 7AB^²