Question

In the given figure, ABCD is a cyclic quadrilateral in which DC is produced to E and CF is drawn parallel to AB such that $\angle ADC={95}^{\circ }and\angle ECF={20}^{\circ }Then,\angle BAD=?$.

(a) ${95}^{\circ }$

(b) ${85}^{\circ }$

(c) ${105}^{\circ }$

(d) ${75}^{\circ }$

Answer 1

(c) 105°
We have:
$\angle ABC$ + $\angle ADC$ = 180°
⇒ $\angle ABC$ + 95° = 180°
⇒$\angle ABC$ = (180° - 95°) = 85°
Now, CF || AB and CB is the transversal.
∴ $\angle BCF$ = $\angle ABC$ = 85° (Alternate interior angles)
⇒ $\angle BCE$ = (85° + 20°) = 105°
⇒ $\angle DCB$ = (180° - 105°) = 75°
Now, $\angle BAD$ + $\angle BCD$ = 180°
⇒ $\angle BAD$ + 75° = 180°
⇒ $\angle BAD$ = (180° - 75°) = 105°