Question

ABCD is a cyclic quadrilateral. Then which of the given angles add to ${180}^{\circ }$?

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is C 3

In a cyclic quadrilateral sum of opposite internal angles is ${180}^{\circ }$.

Therefore opposite internal angles in a cyclic quadrilateral are supplementary.

Let O be the centre of the circle that can pass through ABCD. AC is the chord of the circle.

Join AC and BD.
Since sum of all the angles in a triangle is ${180}^{\circ }$,
we have that $\angle$CAB + $\angle$ABC + $\angle$CBA = ${180}^{\circ }$
Also angles in a same segment are equal we have that $\angle$CAB = $\angle$BDCand also $\angle$ACB = $\angle$ADB.
Adding both the equations we get $\angle$BDC + $\angle$ADB = $\angle$CAB + $\angle$ACB.
If we add$\angle$ABC on both sides, we get that$\angle$ABC + $\angle$BAC + $\angle$ACB = $\angle$ABC + $\angle$ADC.
As $\angle$ABC, $\angle$BAC, $\angle$ACB are angles in a triangle, their sum will be ${180}^{\circ }$.
Therefore $\angle$ABC + $\angle$ADC = ${180}^{\circ }$.

Similarly we can prove that other pair of internal opposite angles are supplementary.
Thus, $\angle$BAD + $\angle$BCD = ${180}^{\circ }$