in an equilateral triangle ABC.D is a point on side BC such that BD=1/3BC.prove that 9AD2=7AB2

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=1/3BC.prove that 9AD2=7AB2
Answer» Given :- An equilateral triangle ABC and D be a point on BC such that BD= 1/3 BC.To prove :- 9AD^2 = 7AB^2Construction :- Draw AE Perpendicular to BC and Join AD.Proof :- Triangle ABC is an equilateral triangle and AE Perpendicular to BCTherefore BE = ECThus, we haveBD = 1/3 BC and DC = 2/3 BC and BE = EC = 1/3BCIn Triangle AEB AE^2 + BE^2 = AB^2 [Using Pythagoras Theorem]AE^2 = AB^2 - BE^2AD^2 - DE^2 = AB^2 - BE^2 [\'.\' In Triangle AED, AD^2 = AE^2 + DE^2]AD^2 = AB^2 - BE^2 + DE^2 AD^2 = AB^2 - (1/2 BC)^2 + ( BE - BD )^2AD^2 = AN^2 - 1/4 BC ^2 + (1/2 BC - 1/3 BC )^2AD^2 = AB^2 - 1/4 BC^2 + ( BC/6 )^2 AD^2 = AB^2 - BC^2 (1/4 - 1/36 )AD^2 = AB^2 - BC^2 (8/36)9AD^2 = 9AB^2 - 2BC^29AD^2 = 9AB^2 - 2AB^2 [ \'.\' AB = BC ]9AD^2 = 7 AB^2 i know the shortest trick. i was just knowing how many here know the short ans?????? This is too long No in RD it is too long.don\'t you have any short sol. Given in Rd Sharma Hi siya