Finding the slopes of the sides of a triangle, when only given the slopes of internal angle bisectors

Question

Finding the slopes of internal or external angle bisectors, when given the slopes of the sides of a triangle, it's a simple question, the one you can easily find in any textbook of analytic geometry.

But the reverse problem : finding the slopes of the sides of a triangle, when only given the slopes of internal angle bisectors. Is it possible? Is there a straightforward formula for that?

Answer 1

Yes. The slopes of the internal bisectors intersect at the incentre I. You will thus know the angles AIB, BIC and CIA.

180 = AIB + IAB + IBA

180 = ACB + 2*IAB + 2*IBA

ACB = 2*AIB - 180

From this you can figure out the side slopes.

Answer 2

Thanks, Paul Childs, finally the solution dawned on me!

First step: Knowing the slopes of the bisectors you calculate tan(A/2), tan(B/2), tan(C/2).

Second step: you calculate the slopes of straight lines from the vertices making these angles to the bisectors.

Case closed.