In $\mathrm{△}ABC$ prove that $\left(r_1+r_2\right){\sec }^2\frac{C}{2}=\left(r_2+r_3\right){\sec }^2\frac{A}{2}=\left(r_3+r_1\right){\sec }^2\frac{B}{2}$

$r_1+r_2=4R\cos \frac{C}{2}$$\left[\sin \frac{A}{2}\cos \frac{B}{2}+\cos \frac{A}{2}\sin \frac{B}{2}\right]$

$=4R\cos \frac{C}{2}\sin \left(\frac{A+B}{2}\right)$

$=4R\cos \frac{C}{2}\left(\cos \frac{C}{2}\right)$

$=4R{\cos }^2\frac{C}{2}$

$\left(r_1+r_2\right){\sec }^2\frac{C}{2}=\frac{4R{\cos }^2\frac{C}{2}}{{\cos }^2\frac{C}{2}}$

$\left(r_1+r_2\right){\sec }^2\frac{C}{2}=4R$

similary, we can show that

$\left(r_2+r_3\right){\sec }^2\frac{A}{2}=\left(r_3+r_1\right)$

${\sec }^2\frac{B}{2}=4R$

$\therefore \left(r_1+r_2\right){\sec }^2\frac{C}{2}=\left(r_2+r_3\right){\sec }^2\frac{A}{2}$

$=\left(r_3+r_1\right){\sec }^2\frac{B}{2}$