Question

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

• A. ${60}^{\circ }$
• B. ${80}^{\circ }$
• C. ${50}^{\circ }$
• D. ${40}^{\circ }$

Answer 1

The correct option is A ${60}^{\circ }$

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

$\angle BDC=\angle BAC$
[Angles formed by the same segment are equal]

$\angle BDC={40}^{\circ }$

$\text{In}\Delta BDC,$

$\angle BDC+\angle DBC+\angle BCD={180}^{\circ }$
[Sum of angles in triangle]

${40}^{\circ }+{80}^{\circ }+\angle BCD={180}^{\circ }$

$\angle BCD={180}^{\circ }-{120}^{\circ }$

$\angle BCD={60}^{\circ }$