1. **Problem Statement:** Find the angle $\angle ADC$ given that $\angle ADB = \angle CDB$ (equal chords subtend equal angles at the circumference) and the circle touches quadrilateral $PQRS$ at points $A, B, C, D$. 2. **Key Concept:** Equal chords in a circle subtend equal angles at the circumference on the same side of the chord. Here, $\angle ADB = \angle CDB$ because $AB = BC$ (given $BA = BC$) and $B$ lies on the circle. 3. **Step-by-step solution:** - Since $BA = BC$, chords $AB$ and $BC$ are equal. - Therefore, angles subtended by these chords at the circumference on the same side are equal: $\angle ADB = \angle CDB$. - The quadrilateral $PQRS$ is tangent to the circle at $A, B, C, D$. - Using the tangent-secant theorem, the angle between a tangent and a chord through the point of contact equals the angle in the alternate segment. - From part (a), $\angle PAB = 40^\circ$. - Since $\angle BAD$ is adjacent to $\angle PAB$, and $\angle BAD$ is part of the quadrilateral, we use the fact that the sum of angles around point $A$ on the circle is $180^\circ$. - Using the properties of cyclic quadrilaterals and equal chords, $\angle ADC$ equals $\angle ABC$. - Given the symmetry and equal chords, $\angle ADC = 40^\circ$. 4. **Final answer:** $$\boxed{\angle ADC = 40^\circ}$$