In ΔABC, AD is perpendicular to BC such that AD^2 =BD. CD. prove that ΔABC is right angles at A

In ΔABC, AD is perpendicular to BC such that AD^2 =BD. CD. prove that

1.	In ΔABC, AD is perpendicular to BC such that AD^2 =BD. CD. prove that ΔABC is right angles at A
Answer» AD2\xa0= BD\xa0{tex}\\times{/tex}\xa0CDor,\xa0{tex}\\frac { A D } { C D } = \\frac { B D } { A D }{/tex}Therefore,\xa0{tex}\\triangle A D C \\sim \\triangle B D A{/tex}\xa0(by SAS)or,\xa0{tex}\\angle{/tex}BAD =\xa0{tex}\\angle{/tex}ACD;{tex}\\angle{/tex}DAC =\xa0{tex}\\angle{/tex}DBA\xa0(Corresponding angles of similar triangles){tex}\\angle{/tex}BAD\xa0+ {tex}\\angle{/tex}ACD + {tex}\\angle{/tex}DAC + {tex}\\angle{/tex}DBA = 180o [sum of angles of ∆]or, 2{tex}\\angle{/tex}BAD + 2{tex}\\angle{/tex}DAC = 180oor,\xa0{tex}\\angle{/tex}BAD + {tex}\\angle{/tex}DAC = 90oTherefore,\xa0{tex}\\angle{/tex}A = 90o