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

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 9ADAD =7AB*AB
Answer» We have {tex}BD = \\frac{1}{3}BC{/tex} Draw AP \u200b{tex} \\bot {/tex}\u200b BCIn \u200b{tex}\\triangle {/tex}\u200bAPB,AB2 = AP2 + BP2 = AP2 + (BD + DP)2= AP2 + BD2 + DP2 + 2BD.DP= (AP2 + DP2) + BD2 + 2BD.DP=AD2 + DB2 + 2BD.DP [AP2 + DP2 = AD2]=AD2+{tex}{\\left( {\\frac{1}{3}BC} \\right)^2} + 2\\left( {\\frac{1}{3}BC} \\right)(BP - BD){/tex}={tex}AD^2 + \\frac{1}{9}BC + \\frac{2}{3}BC\\left( {\\frac{1}{2}BC - \\frac{1}{3}BC} \\right){/tex}{tex}\\left[ {BP = \\frac{1}{2}BC,BD = \\frac{1}{3}BC} \\right]{/tex}={tex}A{D^2} + \\frac{1}{9}A{B^2} + \\frac{2}{3}AB\\left( {\\frac{1}{2}AB - \\frac{1}{3}AB} \\right){/tex} [BC = AB, Sides of an equilateral{tex}\\vartriangle {/tex}\u200b ]={tex}A{D^2} + \\frac{1}{9}A{B^2} + \\frac{1}{3}A{B^2} - \\frac{2}{9}A{B^2}{/tex}={tex}A{D^2} + \\frac{1}{9}A{B^2} + \\frac{1}{9}A{B^2}{/tex}\u200b{tex}\\Rightarrow {/tex}\u200b {tex}A{B^2} = A{D^2} + \\frac{2}{9}A{B^2}{/tex}\u200b{tex}\\Rightarrow {/tex}\u200b 9AB2 = 9AD2 + 2AB2\u200b{tex}\\Rightarrow {/tex}\u200b9AB2 = 9AB2 + 2AD2\u200b{tex}\\Rightarrow {/tex}\u200b 7AB2 = 9AD2\u200b{tex}\\Rightarrow {/tex}\u200b 9AD2 = 7AB2 Proved