Chapter 7.3, Problem 102E

To determine

To show: that the length of $s$ is given by $s=\frac{2ab\cos x}{a+b}$ .

Explanation of Solution

Given information:

In triangle $ABC$ the line segment $s$ bisects angle $C$ .

Given triangle,

  Calculation:

Consider the following sketch,

  

Need to prove the following $s=\frac{2ab\cos x}{a+b}$ using the Law of sines,

  $\begin{array}{l}{A}_1=\frac{1}{2}\left(a\right)\left(s\right)\sin x\\ A=\frac{1}{2}\left(a\right)\left(b\right)\sin c\\ A_2=\frac{1}{2}\left(b\right)\left(s\right)\sin x\\ A=\frac{1}{2}\left(a\right)\left(b\right)\sin c\\ A=\frac{1}{2}\left(a\right)\left(b\right)\sin 2x\end{array}$

Finally set the total area of the triangle equal to the sum of the smaller two,

  $\begin{array}{c}A=A_1+A_2\\ \frac{1}{2}ab\sin 2x=\frac{1}{2}as\sin x+\frac{1}{2}bs\sin x\\ ab\sin 2x=as\sin x+bs\sin x\\ 2ab\sin x\cos x=as\sin x+bs\sin x\\ 2ab\sin x\cos x=s\sin x\left(a+b\right)\\ 2ab\cos x=s\left(a+b\right)\\ s=\frac{2ab\cos x}{a+b}\end{array}$