Sides AB and DC of cyclic quadrilateral ABCD are produced to meet at E, and sides AD and BC are produced to meet at F. If $\angle ADC = 75^\circ$ and $\angle BEC = 52^\circ$ then the difference between $\angle BAD$ and $\angle AFB$ is:

$31^\circ$

\(\angle\)ADC = \({75}^\circ\) and \(\angle\)BEC = \({52}^\circ\)

As we know, in a cyclic quadrilateral, the sum of opposite angles are \({180}^\circ\).

\(\angle\)ADC + \(\angle\)ABC = \({180}^\circ\)

= \(\angle\)ABC = 180 - 75 = 105

= \(\angle\)ABC + \(\angle\)CBE = 180 [straight line]

= \(\angle\)CBE = 180 - 105 = 75

In \(\Delta \)BEC

\(\angle\)CBE + \(\angle\)BEC + \(\angle\)ECB = 180

= \(\angle\)ECB = 180 - 75 - 52 = 53

= \(\angle\)ECB + \(\angle\)BCD = 180 [straight line]

= \(\angle\)BCD = 180 - 53 = 127

= \(\angle\)BAD + \(\angle\)BCD = 180

= \(\angle\)BAD = 180 - 127 = 53

In \(\Delta \)AFB

\(\angle\)BAF + \(\angle\)ABF + \(\angle\)AFB = 180

= \(\angle\)AFB = 180 - 53 - 105 = 22

Therefore, difference between \(\angle\)BAD and \(\angle\)AFB = \({53}^\circ\) - \({22}^\circ\) = \({31}^\circ\).