Question

∆ABC,AD perpendicular to BC and AD square=BD×DC prove angle BAC=90

Answer 1

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

To prove : BAC = 90°

Proof : in right triangles ∆ADB and ∆ADC
So, Pythagoras theorem should be apply ,
Then we have ,
AB² = AD² + BD² ----------(1)
AC²= AD²+ DC² ---------(2)

AB² + AC² = 2AD² + BD²+ DC²
= 2BD . CD + BD² + CD² [ ∵ given AD² = BD.CD ]
= (BD + CD )² = BC²
Thus in triangle ABC we have , AB² + AC²= BC²

hence triangle ABC is a right triangle right angled at A

∠ BAC = 90°