Question

ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If $\angle DBC={80}^{\circ }$ and $\angle BAC$ is ${40}^{\circ }.$ Find $\angle BCD$.

Answer 1

Given, $\angle BAC={40}^{\circ }$ and $\angle DBC={80}^{\circ }$

$\angle BDC=\angle BAC$ [Angles in the the same segment are equal]
So, $\angle BDC={40}^{\circ }$

In $\Delta BDC,$
$\angle BDC+\angle DBC+\angle BCD={180}^{\circ }$ [Sum of angles in triangle]
${40}^{\circ }+{80}^{\circ }+\angle BCD={180}^{\circ }$
$\angle BCD={180}^{\circ }-{120}^{\circ }$
$\angle BCD={60}^{\circ }$