Question

In the given figure, O is the centre of the circle passing through the point A,B,C and D and DC is produced to point E. If angle BAD = 60 find angle BCE and angle BOD

Answer 1

$\angle BCE,\angle BOD=?$
$\angle BOD=2\angle DAB$ (Angle at centre is twice)
$\angle BOD=2\times {60}^{\circ }={120}^{\circ }$
ABCD is cyclic quadrilateral
$\angle DAB+\angle BCD={180}^{\circ }$
$\angle BCD={180}^{\circ }-{60}^{\circ }={120}^{\circ }$
$\angle BCE={180}^{\circ }-{120}^{\circ }={60}^{\circ }$