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.

  • $\begingroup$ Start by drawing a diagram. Also note that $\dfrac {2F}c$ is a height of the triangle. $\endgroup$
    – player3236
    Nov 23 '20 at 12:48

enter image description here

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.


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