Quadrilateral

Let the point $D$ falls on the circle be $D'$ . Then $ABCD'$ is a cyclic quadrilateral and $\angle AD'C + \angle ABC = 180^\circ$ . $\implies \angle AD'C =$ $180^\circ - \angle ABC =$ $180^\circ - 120^\circ =$ $60^\circ$ . Then the original $\angle ADC =$ $\dfrac {\angle AD'C}3 =$ $\dfrac {60^\circ}3 =$ $20^\circ$ .

Now, let the orginal $\angle DCB = \theta$ . Since the sum of internal angles is $360^\circ$ , $\angle DAB =$ $360^\circ - \angle ABC - \angle BCD - \angle ADC =$ $360^\circ - 120^\circ - \theta - 20^\circ =$ $220^\circ - \theta$ . When $\angle ADC$ decreases to $\angle AD'C$ by $x^\circ$ . Then $\angle AD'C + \angle DAB =$ $\theta - x + 220^\circ - \theta =$ $180^\circ$ $\implies x =$ $\boxed{40}^\circ$ .

for quadrilateral inscribed inside a circle, opposite angles sum up to 180 degree. So, final value of < ADC is 60 degree and so, initial <ADC is 20 degree. Now, initial <ADC + x = final <ADC. so, x= 40.