ABC is a triangle in which AB = AC and D is any point on BC prove that

1.	ABC is a triangle in which AB = AC and D is any point on BC prove that AB^2-AD^2=BD×CD
Answer» Draw {tex}A E \\perp B C{/tex}In\xa0{tex}\\Delta{/tex}AEB and {tex}\\Delta{/tex}AEC, we haveAB = AC,AE = AE [Common]and, {tex}\\angle{/tex}B = {tex}\\angle{/tex}C [{tex}\\because{/tex}\xa0AB = AC]{tex}\\therefore \\quad \\Delta A E B \\cong \\Delta A E C{/tex}{tex}\\Rightarrow{/tex}\xa0BE = CE [by CPCT]Since {tex}\\Delta{/tex}AED and {tex}\\Delta{/tex}ABE are right triangles right-angled at E.Therefore,{tex}\\Rightarrow{/tex}\xa0AD2 = AE2 + DE2 and AB2 = AE2 + BE2\xa0{tex}\\Rightarrow{/tex}\xa0AB2 -\xa0AD2 = BE2\xa0- DE2{tex}\\Rightarrow{/tex}\xa0AB2 - AD2 = (BE+ DE) (BE - DE){tex}\\Rightarrow{/tex}\xa0AB2 - AD2 = (CE+ DE) (BE - DE) [{tex}\\because{/tex}\xa0BE = CE ]{tex}\\Rightarrow{/tex}\xa0AB2- AD2\xa0= CD·BDHence, AB2 - AD2 = BD·CD

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