Question

A quadrilateral is inscribed in a circle. If angles are inscribed in the four ares cut off by sides of the quadrilateral, without intersecting the sides between vertices, their sum will be :

• A. ${180}^{\circ }$
• B. ${540}^{\circ }$
• C. ${360}^{\circ }$
• D. ${450}^{\circ }$
• E. ${1080}^{\circ }$

Answer 1

The correct option is B ${540}^{\circ }$

ABCD be quadrilateral.

O is the center of the circumcircle.

$A_1$, $B_1$, $C_1$, $D_1$ be the points on the arcs which make angles by joining vertices without cutting sides.

M be the point such that $\angle AMD$ is the angle made by $AD$ opposite to point $A_1$

Let $\angle AOD=2\theta$

$\therefore \angle AMD=\theta$

$AMD{A}_1$ is a cyclic quadrilateral.

$\therefore \angle A{A}_{1}D+\angle AMD=180°\therefore \angle A{A}_{1}D=180°-\theta$

Sum of angles at the center of the circle made by the sides = 360°

$\sum 2\theta =360°\sum \theta =180°$

$\therefore$ Sum of the angles on arcs = $\sum (180°-\theta )=4\left(180°\right)-\sum \theta =720°-180°=540°$