in triangle ABC,if AD is perpendicular to BCand AD*2=BD.DC,prove that angle BAC=90 degree

 Given : In triangle ABC , AD is perpendicular to BC & AD² = BD.DC

To proove : BAC = 90 degree

Proof : in right triangles ADB & ADC we have AB² = AD² +BD² . . . . . . . . .1

& AC² = AD²+ DC² . . . . . . . .2

from 1 & 2 , AB² + AC² = 2AD² . BD² + DC²      

                                          = 2BD . CD + BD² + CD^{2         { AD*2 = BD . CD (GIVEN ) }}

                                          = (BD + CD )² = BC²

Thus in triangle ABC we have , AB² + AC² = BC²

hence triangle ABC is a right triangle right angled at  A

so BAC = 90 degree      ( proved )