We know that,
$$
\begin{aligned}
& r=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \\
& r_1=4 R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \\
& r_2=4 R \sin \frac{B}{2} \cos \frac{A}{2} \cos \frac{C}{2} \\
& r_3=4 R \sin \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2}
\end{aligned}
$$
Given that,
$$
\begin{aligned}
& r_3=r_1+r_2+r \\
& \Rightarrow \quad r_3-r=r_1+r_2 \\
& \Rightarrow \quad 4 R \sin \frac{C}{2}\left(\cos \frac{A}{2} \cos \frac{B}{2}-\sin \frac{B}{2} \sin \frac{A}{2}\right) \\
& =4 R \cos \frac{C}{2}\left[\sin \frac{A}{2} \cos \frac{B}{2}+\cos \frac{A}{2} \sin \frac{B}{2}\right] \\
& \Rightarrow \quad \sin \frac{C}{2}\left(\cos \left(\frac{A+B}{2}\right)\right)=\cos \frac{C}{2}\left(\sin \left(\frac{A+B}{2}\right)\right) \\
& \Rightarrow \sin \frac{C}{2}\left(\cos \left(\frac{\pi}{2}-\frac{C}{2}\right)\right)=\cos \frac{C}{2}\left(\sin \left(\frac{\pi}{2}-\frac{C}{2}\right)\right) \\
& {\left[\because A+B+C=\pi \Rightarrow \frac{\mathrm{A}}{2}+\frac{B}{2}=\frac{\pi}{2}-C\right]} \\
& \Rightarrow \quad \sin ^2 \frac{A}{2}=\cos ^2 \frac{C}{2} \\
& \Rightarrow \quad \tan \frac{C}{2}=1 \\
&
\end{aligned}
$$
$$
\Rightarrow \quad \frac{C}{2}=\frac{\pi}{4} \Rightarrow C=\frac{\pi}{2}
$$
We know that, $A+B+C=\pi$
$$
\begin{array}{ll}
\Rightarrow & A+B=\pi-\frac{\pi}{2} \\
\Rightarrow & A+B=\frac{\pi}{2}
\end{array}
$$