Question

ABCD is a cyclic quadrilateral, in which \(\angle DBC = 80^\circ\) and \(\angle BAC = 40^\circ\). __ is the $\angle BCD$.

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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 = ${180}^{\circ }$ - ${120}^{\circ }$ = ${60}^{\circ }$