Let $\angle{ADC}=x$. It is easy to find $\angle{ADC} = x+43^\circ$. Apply the Ceva's theorem (in the form of angles) on point $A$ against $\triangle{BDC}$: $$1=\frac{\sin\angle{DCA}}{\sin\angle{ACB}}\cdot\frac{\sin\angle{CBA}}{\sin\angle{ABD}}\frac{\sin\angle{BDA}}{\sin\angle{ADC}}=\frac{\sin{13^\circ}}{\sin{73^\circ}}\cdot\frac{\sin{77^\circ}}{\sin{26^\circ}}\cdot\frac{\sin{x}}{\sin{(x+43^\circ)}}$$ It follows that: \begin{align*} \frac{\sin(x+43^\circ)}{\sin{x}} &= \frac{\sin{13^\circ}}{\sin{73^\circ}}\cdot\frac{\sin{77^\circ}}{\sin{26^\circ}}\\ &=\frac{\sin{13^\circ}}{\sin{73^\circ}}\cdot\frac{\cos{13^\circ}}{\sin{26^\circ}}\\ &=\frac{1}{2\sin{73^\circ}}\\ &=\frac{\sin{30^\circ}}{\sin{73^\circ}}\\ &=\frac{\sin{150^\circ}}{\sin{107^\circ}}\\ &=\frac{\sin(107^\circ + 43^\circ)}{\sin{107^\circ}} \end{align*} Therefore, we conclude $\angle{ADB}=x=107^\circ$.