Question

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

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

Answer 1

The correct option is A

${60}^{\circ }$

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

Since the angles formed by the same segment are equal,
$\angle BDC=\angle BAC={40}^{\circ }.$

$\text{In}\Delta BDC,$

$\angle BDC+\angle DBC+\angle BCD={180}^{\circ }.\text{[Angle sum property]}$

i.e., ${40}^{\circ }+{80}^{\circ }+\angle BCD={180}^{\circ }$

$⟹\angle BCD={180}^{\circ }-{120}^{\circ }={60}^{\circ }$