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 $\triangle ABC.$

he segments $BB'$ and $A'C'$ meet at point $D.$ Find $\frac {BI}{BB'}, \frac {DA'}{DC'}, \frac {BD}{BB'}.$

Solution

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

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

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

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

Bisector $BD = 2 \frac {BC' \cdot BA'}{BC' + BA'} \cos \beta \implies$ $\frac {BD}{BB'} = \frac{a+c}{a+2b+c}.$ vladimir.shelomovskii@gmail.com, vvsss

Proportions for bisectors

The bisectors $AE$ and $CD$ of a triangle ABC with $\angle B = 60^\circ$ meet at point $I.$

Prove $\frac {CD}{AE} = \frac {BC}{AB}, DI = IE.$

Proof

Denote the angles $A = 2\alpha, B = 2\beta = 60^\circ, C = 2 \gamma.$ $\angle AIC = 180^\circ - \alpha - \gamma = 90^\circ + \beta = 120^\circ \implies B, D, I,$ and $E$ are concyclic. $\angle BEA = \angle BEI = \angle ADC.$ The area of the $\triangle ABC$ is $[ABC] = AB \cdot h_C = AB \cdot CD \cdot \sin \angle ADC = BC \cdot AE \cdot \sin \angle AEB \implies$ $\frac {CD}{AE} = \frac {BC}{AB} = \frac {a}{c}.$ $\frac {DI}{IE} = \frac {DI}{CD} \cdot \frac {AE}{IE}\cdot \frac {CD}{AE}= \frac {c}{a+b+c} \cdot \frac {a+b+c} {a} \cdot \frac {a}{c} = 1.$ vladimir.shelomovskii@gmail.com, vvsss

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Bisector

Division of bisector

Proportions for bisectors