In the given figure, O is the centre of circle. The angle subtended by the arc BCD at the centre is 140°. BC is produced to P. Determine ∠BAD + ∠DCP.

∠BOD = 140°

$\therefore \angle \mathrm{BAD}=\frac{1}{2} \angle \mathrm{BOD}=70^{\circ}$

$\mathrm{ABCD}$ is a concyclic quadrilateral.

$\therefore \angle \mathrm{BAD}+\angle \mathrm{BCD}=180^{\circ}$

$\Rightarrow 70^{\circ}+\angle \mathrm{BCD}=180^{\circ}$

$\Rightarrow \angle \mathrm{BCD}=110^{\circ}$

$\therefore \angle \mathrm{BCD}+\angle \mathrm{DCP}=180^{\circ}$

$\Rightarrow 110^{\circ}+\angle \mathrm{DCP}=180^{\circ}$

$\Rightarrow \angle \mathrm{DCP}=180^{\circ}-110^{\circ}=70^{\circ}$

$\therefore \angle \mathrm{BAD}=\angle \mathrm{DCP}=70^{\circ}$

$\therefore \angle \mathrm{BAD}+\angle \mathrm{DCP}=140^{\circ}$