Length of Angle Bisector

Theorem
Length of bisectors

$$b_\alpha^2=\dfrac{cb}{(c+b)^2}[(c+b)^2-a^2]$$

$$b_\gamma^2=\dfrac{ab}{(a+b)^2}[(a+b)^2-c^2]$$

$$b_\beta^2=\dfrac{ac}{(a+c)^2}[(a+c)^2-b^2]$$

where $$b_\alpha$$,$$b_\gamma$$ and $$b_\beta$$ are bisector from $$A$$,$$C$$ and $$B$$ ,respectively

Proof 1
Using Stewart's Theorem and the Theorem of Internal Bisector then $$\dfrac{AP}{PB}=\dfrac{b}{a}$$ and $$CP=b_\gamma$$.

Now we use

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using that in Stewart's Theorem we have

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A similar argument can be used to show that the statement holds for the others bisector

Proof 2


we have u and v that are the segment that bisector leaves(see figure above)

$$u=\dfrac{bc}{a+b}$$

$$v=\dfrac{ac}{a+b}$$

we have

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Then by AA similarity we have

$$\triangle CAD \sim \triangle CFB$$

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now We use the Chord theorem then we have $$v \cdot u=b_\gamma \cdot DF$$

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