In the following figure,
If $\angle BAD = 96^\circ,$ find $\angle BCD$ and $\angle BFE.$
Exercise 17 (A) | Q 19.1 | Page 259
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$ABCD$ is a cyclic quadrilateral
$ \therefore \angle BAD +\angle BCD =180^{\circ} $
(pair of opposite angles in a cyclic quadrilateral are supplementary)
$ \Rightarrow \angle BCD =180^{\circ}-96^{\circ}=84^{\circ}$
$\therefore \angle BCE =180^{\circ}-84^{\circ}=96^{\circ} $
Similarly, $BCEF$ is a cyclic quadrilateral
$ \therefore \angle B C E+\angle B F E=180^{\circ} $
(pair of opposite angles in a cyclic quadrilateral are supplementary)
$ \therefore \angle B F E=180^{\circ}-96^{\circ}=84^{\circ} $
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In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate : ∠DAB
Also show that the ΔAOD is an equilateral triangle .