hard bisector geometry help

Hi proyaop, yes your answer is correct.

Here is another method, using the cosine law twice only.

Consider triangle ABM, AB=5, AM=4 and BM = 4 (Because the median bisects the opposite side to the vertex).

So applying cosine rule gives you the angle ABC (which is the same as ABM)

Then, consider the triangle ABC, we have, AB=5 and BC=8 and we have the angle ABC.

Then apply cosine rule, yielding AC to be \(\sqrt{39}\)

Thanks for double checking, also I think you can use AC as the diameter of a circle, then knowing ABC is a right triangle and then using the pythagorean theorem, thanks anyway :D

Wow! This is really a neat solution!