Question

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



[ 2 marks ]

Answer 1

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

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

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


​​​​​​​[ 1 mark ]

$\text{In}\Delta BDC,$

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

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

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

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

[ 1 mark ]