Question

AD is an altitude of an isosceles triangle ABC is which AB=AC. prove that ∠BAD =∠DAC.
one to answer it first will be the brainiest.

Answer 1

See the diagram,
Given AB=AC
Let AB=AC=y
suppose angle BAD= Bita
& angle DAC = Alpha

from figure it is clear that for ∆BAD & ∆DAC,
AD are equal,

now applying cos formula for both ∆BAD & ∆DAC,

therefore cosBita = AD/AC = AD/y......(1)
also cosAlpha = AD/AB = AD/y............(2)

comparing (1)&(2) we get,
cosBita = cosAlpha
Bita = Alpha
So,
angle BAD = angle DAC.

_-_-_-_-_-_HENCE PROVED_-_-_-_-_