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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?

MrDudulex
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2 Answers2

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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.

Paul Childs
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  • Could you explain how? For instance, if m and m' are the slopes of BI and CI, I know that tan(BIC)= -|(m'-m)/1+m'm| because BIC is obtuse, but I don't know how to proceed further. – MrDudulex Sep 10 '18 at 13:08
  • If you know the slopes of the equations of the angle bisectors, I know that it's fairly easy to find the angles of the triangle, but finding the slopes of the equations of the sides remains puzzling... – MrDudulex Sep 10 '18 at 13:20
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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.

MrDudulex
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