ABCD is a cyclic quadrilateral. If \(\angle ADB = x^\circ\) and \(\angle ABD = y^\circ\), then the measure of \(\angle BCD\) will be \((x + y)^\circ\).

Q.ABCD is a cyclic quadrilateral. If \(\angle ADB = x^\circ\) and \(\angle ABD = y^\circ\), then the measure of \(\angle BCD\) will be \((x + y)^\circ\).

The statement is true.

In triangle ABD, \(\angle ADB = x^\circ\) and \(\angle ABD = y^\circ\) \[ \therefore \angle BAD = 180^\circ - (\angle ADB + \angle ABD) = 180^\circ - (x^\circ + y^\circ) \quad -----(i) \] Now, since ABCD is a cyclic quadrilateral, \[ \angle BAD + \angle BCD = 180^\circ \Rightarrow \angle BCD = 180^\circ - \angle BAD \quad -----(ii) \Rightarrow \angle BCD = 180^\circ - [180^\circ - (x^\circ + y^\circ)] \therefore \angle BCD = (x + y)^\circ \]