Question

In the figure at left, ABCD is a cyclic quadrilateral in which AB || DC. If $\angle BAD={100}^o$
Find (i)$\angle BCD$
(ii)$\angle ADC$
(iii)$\angle ABC$

Answer 1

(i) $\angle BAD+\angle BCD=180°$ (opposite angles of a cyclic quadrilateral are supplementary).

$\therefore 100°+\angle BCD=180°\angle BCD=80°$

(ii) $\angle ADC=\angle BCD$ (Alt angles are equal)

$\angle ADC=80°$

(iii) $\angle ABC=\angle BAD$ (Alt angles are equal)

$\angle ABC=100°$