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The internal bisector from A cuts side a into pieces proportional to the adjacent sides (BD:DC = c:b) and has length √[bc((b+c)² − a²)] / (b+c).
Bisector length
tₐ = √[bc((b+c)² − a²)] / (b + c)
Angle bisector theorem
BD : DC = c : b
The bisector from a vertex divides the opposite side into two segments whose lengths are in the same ratio as the two sides forming the angle.
Yes, at the incentre, which is the centre of the inscribed circle.