In a circle, ABCD is a cyclic quadrilateral. AC and BD intersect each other at P. If AB = AC and ∠BAC = 48°, then the measure of ∠ADC is

114°

\(\angle\)ABC = \(\angle\)ACB  [as AB = AC]

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

So, \(\angle\)ABC = \(\angle\)ACB = (\({180}^\circ\) - \({48}^\circ\))/2 = \({132}^\circ\)/2 = \({66}^\circ\)

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

\(\angle\)ADC + \({66}^\circ\) = \({180}^\circ\)

\(\angle\)ADC = \({180}^\circ\) - \({66}^\circ\) = \({114}^\circ\)

Therefore, \(\angle\)ADC is \({114}^\circ\).