In a quadrilateral△ABC D is a pount on side BC such that 4BD =BC. prove that 16ADsquare =13BC square

In a quadrilateral△ABC D is a pount on side BC such that 4BD =BC. pr

1.	In a quadrilateral△ABC D is a pount on side BC such that 4BD =BC. prove that 16ADsquare =13BC square
Answer» In equilateral {tex}\\triangle{/tex}ABC. 4BD = BCConstruction: Draw AE\xa0{tex}\\perp{/tex} BC.\xa0{tex}\\therefore{/tex} BE = {tex}\\frac{1}{2}{/tex}BC.In right {tex}\\triangle{/tex}AED, AD2 = DE2 + AE2\xa0{tex}\\Rightarrow{/tex} AE2 = AD2 - DE2 ..(i)In right {tex}\\triangle{/tex}AEB, AB2 = AE2 + BE2{tex}\\Rightarrow{/tex}\xa0AB2 = AD2 - DE2 + BE2 [using (i)]{tex}\\Rightarrow{/tex}\xa0AB2 + DE2 - BE2 = AD2{tex}\\Rightarrow{/tex}\xa0AB2 + (BE - BD)2 - BE2 = AD2{tex}\\Rightarrow{/tex}\xa0AB2\xa0+ BE2\xa0+ BD2\xa0- 2BE.BD - BE2\xa0= AD2{tex}\\Rightarrow{/tex}\xa0AB2 + ({tex}\\frac{1}{2}{/tex}BC)2 - {tex}2 \\times \\frac{1}{2}{/tex}BC{tex}\\times \\frac{1}{4}{/tex}BC = AD2{tex}\\Rightarrow{/tex}\xa0AB2\xa0+\xa0{tex}\\frac{1}{16}{/tex}BC2\xa0-\xa0{tex}\\frac{1}{4}{/tex}BC2\xa0= AD2{tex}\\Rightarrow{/tex}\xa0BC2\xa0- {tex}\\frac{3}{16}{/tex}BC2\xa0= AD2\xa0[{tex}\\because{/tex} AB = BC]{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { 13 \\mathrm { BC } ^ { 2 } } { 16 }{/tex} = AD2{tex}\\Rightarrow{/tex}\xa013BC2\xa0= 16AD2