ABC is an isosceles triangle in which the equal sides are AB and A.If AD is an altitude and |BC|=1/2|AB|.Prove that |AD|^2=15|BC|^2
2 answers
I feel like this question might be missing some vital information. Is this a 3D trigonometry question? If so, I'm not sure it can be solved without being given an additional length or some bearings/elevation angles.
The altitude in an isosceles triangle right-bisects the base, so BD = DC
let BD = x
then BC = 2x
AB = 4x
in the right-triangle ABD
AD^2 + x^2 = (4x)^2 = 16x^2
AD^2 = 15x^2
AD^2 = 15BD^2 , not 15 BC^2 as you stated.
let BD = x
then BC = 2x
AB = 4x
in the right-triangle ABD
AD^2 + x^2 = (4x)^2 = 16x^2
AD^2 = 15x^2
AD^2 = 15BD^2 , not 15 BC^2 as you stated.
1 answer