ABC is a triangle in which AB = AC and D is any point on BC prove that AB^2-AD^2=BD×CD
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
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