Question

In the given figure, ABCD is a cylic quadrilateral in which BC = CD and $\angle CBD={35}^{\circ }.$ Then, $\angle BAD=?$

(a) ${65}^{\circ }$

(b) ${70}^{\circ }$

(c) ${110}^{\circ }$

(d) ${90}^{\circ }$

Answer 1

ANSWER:
( b ) 70°
BC = CD (given)
⇒ $\angle BDC$ = $\angle CBD$ = 35°
In Δ BCD, we have:
$\angle BCD$ + $\angle BDC$ + $\angle CBD$ = 180° (Angle sum property of a triangle)
⇒ $\angle BCD$ + 35° + 35° = 180°
⇒ $\angle BCD$ = (180° - 70°) = 110°
In cyclic quadrilateral ABCD, we have:
$\angle BAD$ + $\angle BCD$ = 180°
⇒ $\angle BAD$ + 110° = 180°
∴ $\angle BAD$ = (180° - 110°) = 70°