In $\mathrm{△}ABC\text{ If }r+r_1+r_2=r_3$ Then $\text{show that ∠}C={90}^{\circ }$

$r_1+r_2=r-r_3\Rightarrow \frac{r_1+r_2}{r-r_3}=1....\left(1\right)$

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

$\cos \frac{A}{2}\sin \frac{B}{2}\cos \frac{C}{2}$

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

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

$=4R\cos \frac{C}{2}\left[\cos \frac{C}{2}\right]$

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

$r-r_3=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$

$-4R\cos \frac{A}{2}\cos \frac{B}{2}\sin \frac{C}{2}$

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

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

$=4R\sin \frac{C}{2}\left[-\sin \frac{C}{2}\right]=4R{\sin }^2\frac{C}{2}$

$\Rightarrow \frac{r-r_3}{r_1+r_2}=\frac{4\cancel{\text{R }}{\sin }^2\frac{C}{2}}{4\cancel{\text{R }}{\cos }^2\frac{C}{2}}={\mathrm{Tan}}^2\frac{C}{2}$

$\therefore {\mathrm{Tan}}^2\frac{\mathrm{C}}{2}=\mathrm{Tan}{45}^{\circ }$

$\frac{\mathrm{C}}{2}={45}^0\therefore \mathrm{C}={90}^{\circ }$