XV OBM

We have $\angle ADB=80^{\circ }$, $\angle ACB=70^{\circ }$. Applying the sine rule repeatedly we have $PD=\frac{\sin 20^{\circ }}{\sin 80^{\circ }}PA$, $PC=\frac{\sin 30^{\circ }}{\sin 70^{\circ }}PB=\frac{\sin 30^{\circ }}{\sin 70^{\circ }}\frac{\sin 30^{\circ }}{\sin 50^{\circ }}PA$. So we have to show that $\sin 30^{\circ }\sin 30^{\circ }\sin 80^{\circ }=\sin 70^{\circ }\sin 50^{\circ }\sin 20^{\circ }$. Indeed we have $\sin 80^{\circ }=2\sin 40^{\circ }\cos 40^{\circ }=4\sin 20^{\circ }\cos 20^{\circ }\cos 40^{\circ }=4\sin 20^{\circ }\sin 70^{\circ }\sin 50^{\circ }$ and $\sin 30^{\circ }\sin 30^{\circ }=\frac{1}{4}$.