Question

ABCD is a cyclic quadrilateral, in which $\angle DBC={80}^{\circ }$ and $\angle BAC={40}^{\circ }$. $\angle BCD$ (in degrees ) = ___ .

Answer 1

To find, $\angle$BCD = ?

Procedure : $\angle$DBC = $\angle$ DAC

(Angles in the same segment)

$\angle$ DAC = ${80}^{\circ }$ ......( $\angle$DBC = ${80}^{\circ }$)

Now, $\angle$DAB = $\angle$DAC + $\angle$CAB

= ${80}^{\circ }$ + ${40}^{\circ }$ = ${120}^{\circ }$

Then, $\angle$DAB + $\angle$BCD = ${180}^{\circ }$

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

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