Bisector

Division of bisector

Let a triangle $\triangle ABC, BC = a, AC = b, AB = c$ be given. Let $AA', BB',$ and $CC'$ be the bisectors of $\triangle ABC.$ The segments $BB'$ and $A'C'$ meet at point $D.$

Find $\frac {BI}{BB'}, \frac {BD}{BB'}, \frac {DA'}{DC'}.$

Solution

$\frac {BA'}{CA'} = \frac {BA}{CA} = \frac {c}{b}, BA' + CA' = BC = a \implies BA' = \frac {a \cdot c}{b+c}.$

Similarly $BC' = \frac {a \cdot c}{a+b}, B'C = \frac {a \cdot b}{a+b}.$ $\frac {BI}{IB'} = \frac {a}{B'C} = \frac{a+c}{b} \implies \frac {BI}{BB'} = \frac {a+c}{a + b +c}.$

Denote $\angle ABC = 2 \beta.$ Bisector $BB' = 2 \frac {a \cdot c}{a + c} \cos \beta.$

Bisector $BD = 2 \frac {BC' \cdot BA'}{BC' + BA'} \cos \beta \implies$ $\frac {BD}{BB'} = \frac{a+c}{a+2b+c}.$

$\frac {DA'}{DC'} = \frac {BA'}{BC'} = \frac {a+ b}{b +c}.$

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Bisector

Division of bisector