# Formula for the angle bisector length with area and two angles

How can I proof $$w_\gamma =\dfrac{2F}{c\cos \left(\frac{\alpha -\beta }{2}\right)}$$ ?

Where $$w_\gamma$$ is the length of the angle bisector of the side $$c$$
and $$\alpha$$, $$\beta$$ are the angles of the vertices $$A$$, $$B$$
and $$F$$ is the area of the triangle.

I have no idea how to start.

• Start by drawing a diagram. Also note that $\dfrac {2F}c$ is a height of the triangle. Nov 23 '20 at 12:48

Without loss of generality, let's assume that $$|AC|>|BC|$$. Then

\begin{align} \triangle CDH_c:\quad w_{\gamma}=|CD| &= \frac{|CH_c|}{\cos\angle H_cCD} \tag{1}\label{1} \\ &=\frac{|CH_c|}{\cos(\tfrac12\gamma-(90^\circ-\beta))} \tag{2}\label{2} \\ &=\frac{|CH_c|}{\cos(\tfrac12(180^\circ-\alpha-\beta)-(90^\circ-\beta))} \tag{3}\label{3} \\ &=\frac{|CH_c|}{\cos\tfrac{\beta-\alpha}2} =\frac{|CH_c|}{\cos\tfrac{\alpha-\beta}2} \tag{4}\label{4} . \end{align}

As @player3236 pointed out in a comment,

\begin{align} |CH_c|&=\frac{2F}c \tag{5}\label{5} . \end{align}

Substitution of \eqref{5} into \eqref{4} provides the desired result

\begin{align} w_{\gamma}&=|CD| =\frac{2F}{c\cdot\cos\tfrac{\alpha-\beta}2} \tag{6}\label{6} . \end{align}

Note that since $$\cos\tfrac{\alpha-\beta}2=\cos\tfrac{\beta-\alpha}2$$, the expression \eqref{6} holds also if $$|AC|<|BC|$$, and it also holds if $$|AC|=|BC|$$ too.