A straightforward trigonometry solution: It suffices to show $$\frac{AD}{BC}=\frac{AD+AB}{AD}\qquad(1)$$By easy angle chasing and sine law we find $$\frac{AD}{BC}=\frac{\sin 30^{\circ}}{\sin 10^{\circ}}$$$$\frac{AD+AB}{AD}=1+\frac{AB}{AD}=1+\frac{\sin70^{\circ}}{\sin 10^{\circ}}$$Combining these and rearranging, $(1)$ becomes $$\sin^230^{\circ}=\sin 10^{\circ}(\sin 30^{\circ}+\sin 70^{\circ})\qquad(2)$$Using sum to product, the right hand side is $$2\sin 10^{\circ}\cos 20^{\circ}\cos 40^{\circ}$$So $(2)$ becomes $$\frac{1}{8}=\sin 10^{\circ}\cos20^{\circ}\cos40^{\circ}$$which is very well-known and easy to prove(mutliply both sides by $\cos 10^{\circ}$ and use double angle formula). This completes the proof. $\noindent\rule{15.5cm}{0.4pt}$ The official solution involves some crazy extra line construction, good luck on finding a synthetic solution.

Let $AB$ intersects $DC$ at $E$ It's well-known that $AD=BE$ So $AD^2=AD\times BE=AE\times BC=BE\times BC+AB\times BC=(AD+AB)BC$ as desired