If bac is 90degree AD is it\'s bisector. If DE is perpendicular to AC. Prove that DE*(AB+AC)=AB*AC

If bac is 90degree AD is it\'s bisector. If DE is perpendicular to AC.

1.	If bac is 90degree AD is it\'s bisector. If DE is perpendicular to AC. Prove that DE(AB+AC)=ABAC
Answer» To prove the given result,we will use the following theorm.The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angleSince AD is the bisector of {tex}\\angle{/tex}A of {tex}\\Delta{/tex}ABC.{tex}\\therefore \\quad \\frac { A B } { A C } = \\frac { B D } { D C }{/tex}\xa0[by above theorm]{tex}\\Rightarrow \\quad \\frac { A B } { A C } + 1 = \\frac { B D } { D C } + 1{/tex}[Adding 1 on both sides]{tex}\\Rightarrow \\quad \\frac { A B + A C } { A C } = \\frac { B D + D C } { D C }{/tex}{tex}\\Rightarrow \\quad \\frac { A B + A C } { A C } = \\frac { B C } { D C }{/tex}\xa0... (i)In {tex}\\Delta{/tex}\'s CDE and CBA, we have{tex}\\angle{/tex}DCE = {tex}\\angle{/tex}BCA = {tex}\\angle{/tex}C [Common]{tex}\\angle{/tex}BAC = {tex}\\angle{/tex}DEC [Each equal to 90°]So, by AA-criterion of similarity, we have{tex}\\Delta{/tex}CDE ~ {tex}\\Delta{/tex}CBA{tex}\\Rightarrow \\quad \\frac { C D } { C B } = \\frac { D E } { B A }{/tex}{tex}\\Rightarrow \\quad \\frac { A B } { D E } = \\frac { B C } { D C }{/tex}\xa0...(ii)From (i) and (ii), we obtain{tex}\\frac { A B + A C } { A C } = \\frac { A B } { D E } \\Rightarrow D E \\times ( A B + A C ) = A B \\times A C{/tex}