in triangle AbC ad is perpendicular to BC such that AD square =BD×DC prove that ∆ABC is a ryt angled at A
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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 + CD2 { 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
Hope it helps u
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 + CD2 { 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
Hope it helps u
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