Question

In the given figure, ABCD is a cyclic quadrilateral in which AB||DC and $\angle BAD={100}^{\circ }.Then,\angle ABC=?$

(a) ${80}^{\circ }$

(b) ${100}^{\circ }$

(c) ${50}^{\circ }$

(d) ${40}^{\circ }$

Answer 1

ANSWER:
( b ) 100°
Since ABCD is a cyclic quadrilateral, we have:
$\angle BAD$ + $\angle BCD$ = 180° (Opposite angles of a cyclic quadrilateral)
⇒ 100 ° + $\angle BCD$ = 180°
⇒ $\angle BCD$ = (180° - 100°) = 80°
Now, AB || DC and CB is the transversal.
∴ $\angle ABC$ + $\angle BCD$ = 180°
⇒ $\angle ABC$ + 80° = 180°
⇒ $\angle ABC$ = (180° - 80°) = 100°