Question

In the figure, ABCD is a cyclic quadrilateral in which $\angle BAD={75}^o$, $\angle ABD={58}^o$ and $\angle ADC={77}^o$, AC and BD intersect at P. Then, find $\angle DPC$.

Answer 1

ABCD is a cyclic quadrilateral.

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

$\Rightarrow {75}^o+\angle BCD={180}^o$

$\Rightarrow \angle BCD={180}^o-{75}^o={105}^o$

and $\angle ADC+\angle ABC={180}^o$

$\Rightarrow {77}^o+\angle ABC={180}^o$

$\Rightarrow ABC={180}^o-{77}^o={103}^o$

$\therefore \angle DBC=\angle ABC-\angle ABD$

$={103}^o-{58}^o={45}^o$

$\because$ Arc AD subtends $\angle ABDand\angle ACD$ in the same segment of the circle.

$\therefore \angle ABD=\angle ACD={58}^o$

$\therefore \angle ACB=\angle BCD-\angle ACD$$={105}^o-{58}^o={47}^o$

Now in $\Delta PBC$, Ext. $\angle DPC=\angle PBC+\angle PCB$

$=\angle DBC+\angle ACB$$={45}^o+{47}^o={92}^o$

Hence, $\angle DPC={92}^o$