An isosceles obtuse triangle

Since $\triangle ABC$ is isosceles with $\angle A = 135^\circ$ , then $\angle B = \angle C = 22.5^\circ$ . Let $AE$ be the altitude of $\triangle ABC$ which passes through center $O$ of the incircle. Then $OB$ bisects $\angle B$ and $\angle OBE = \angle OBA = 11.25^\circ$ .

We have $BE = \dfrac {10}{\tan 11.25^\circ}$ and $AB = \dfrac {BE}{\sin 22.5^\circ} = \dfrac {10}{\tan 11.25^\circ \sin 22.5^\circ} \approx 54.41553054$ .

By sine rule we have $\dfrac {AB}{\sin \angle ADB} = \dfrac {AD}{\sin \angle ABD} = \dfrac {BD}{\sin \angle BAD}$ , then

$\begin{aligned} AD & = \frac {AB \cdot \sin 11.25^\circ}{\sin 33.75^\circ} \approx 19.10819325 \\ BD & = \frac {AB \cdot \sin 135^\circ}{\sin 33.75^\circ} \approx 69.25783342 \\ \implies AD+BD & \approx \boxed{88.4} \end{aligned}$