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Division of an Angle Bisector by the Incenter

​
AI
≈
8.67
IA'
≈
2.66
AB
≈
11.69
AC
≈
12.13
BC
≈
7.31
AI
IA'
≈
3.26
AB+AC
BC
≈
3.26
Let ABC be a triangle with incenter I. Let AA' be the bisector of
∠BAC
with A' on BC. Then
AI
IA'
=
AB+AC
BC
.
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