Question

In the figure at right, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that $\angle DBC={30}^o$ and $\angle BAC={50}^o$.
Find (i) $\angle BCD$ (ii) $\angle CAD$

Answer 1

Consider $\angle CAD$ and $\angle CBD$

They are both angles made by the same chord $CD$

$\Rightarrow \angle CAD=\angle CBD={30}^{\circ }$

Since $ABCD$ is a cyclic quadrilateral opposite angles are supplementary.

$\Rightarrow \angle DAB+\angle DCB={180}^{\circ }$

$\angle DCB={180}^{\circ }–(\angle DAC+\angle CAB)$

$\angle DCB={180}^o–{30}^o–{50}^o={100}^{\circ }$