For the angle bisector I a </msub> in a triangle A B C it holds

arridsd9 2022-06-25 Answered
For the angle bisector I a in a triangle A B C it holds
I a 2 = b c ( b + c ) 2 [ ( b + c ) 2 a 2 ]
If I is the incenter, I wonder if there exist similar formula for the part A I 2 .
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Answers (2)

Marlee Norman
Answered 2022-06-26 Author has 18 answers
Put weights a, b, c on the three vertices A, B, C. Their center of gravity is I. The center of gravity of B and C is the intersection of A I with B C.
Therefore, A I : I a = b + c a + b + c

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Roland Waters
Answered 2022-06-27 Author has 6 answers
Thank you very much. So the formula is
A I 2 = b c ( a + b + c ) 2 [ ( b + c ) 2 a 2 ] .

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