In an eqilateral triangle ABCD is a point on side BC such that BD=1/3BC prove that 9AD2=7AB2

In an eqilateral triangle ABCD is a point on side BC such that BD=1/3B

1.	In an eqilateral triangle ABCD is a point on side BC such that BD=1/3BC prove that 9AD2=7AB2
Answer» Given, {tex}\\triangle ABC{/tex}\xa0in which\xa0{tex}AB = BC = CA{/tex} and D is a point on BC such that BD ={tex}\\frac { 1 } { 3 }{/tex}BC.Prove: 9AD2 = 7AB2.Construction :Draw\xa0{tex}A L \\perp B C{/tex}\xa0Proof: In right triangles ALB and ALC, we have{tex}AB = AC {/tex}(given){tex}AL = AL {/tex}(common){tex}\\therefore \\triangle A L B \\cong \\triangle A L C{/tex}\xa0[by RHS axiom]So,\xa0{tex}BL = CL.{/tex}Thus, BD ={tex}\\frac { 1 } { 3 }{/tex}BC and BL ={tex}\\frac { 1 } { 2 }{/tex}BC.In\xa0{tex}\\triangle ALB{/tex},\xa0{tex}\\angle A L B = 90 ^ { \\circ }{/tex}By using pythagoras theorem, we get\xa0{tex}\\therefore{/tex}\xa0AB2 = AL2 + BL2 ...(i)In\xa0{tex}\\triangle ALD{/tex},\xa0{tex}\\angle A L D = 90 ^ { \\circ }{/tex}By using pythagoras theorem, we get\xa0{tex}\\therefore{/tex}\xa0AD2 = AL2 + DL2\xa0= AL2 + (BL - BD)2= AL2 + BL2 + BD2 -\xa0{tex}2 B L \\cdot B D{/tex}= (AL2 + BL2) + BD2 -\xa0{tex}2 B L \\cdot B D{/tex}= AB2 + BD2 -\xa0{tex}2 B L \\cdot B D{/tex}\xa0[using (i)]{tex}= B C ^ { 2 } + \\left( \\frac { 1 } { 3 } B C \\right) ^ { 2 } - 2 \\left( \\frac { 1 } { 2 } B C \\right) \\cdot \\frac { 1 } { 3 } B C{/tex}{tex}= B C ^ { 2 } + \\frac { 1 } { 9 } B C ^ { 2 } - \\frac { 1 } { 3 } B C ^ { 2 }{/tex}{tex}= \\frac { 7 } { 9 } B C ^ { 2 } = \\frac { 7 } { 9 } A B ^ { 2 }{/tex}\xa0[{tex}\\because{/tex}\xa0BC = AB]Therefore, 9AD2 = 7AB2.