In the adjoining figure, AOB is a diameter of the circle with center O. Point M is the mid-point of chord CD. $$\angle CDO = 32^\circ$$, $$\angle DBE = 35^\circ$$ and $$\angle BEO = x^\circ$$, then $$x =$$

Since $$OC = OD$$, triangle OCD is isosceles, so $$\angle OCD = \angle ODC = 32^\circ$$ and $$\angle COD = 116^\circ$$. The mid-point M of chord CD lies on the diameter AB, so AB is perpendicular to CD and C, D are symmetric about AB, making arc AC and arc AD each $$58^\circ$$. The inscribed angle $$\angle DBE = 35^\circ$$ gives arc DE $$= 70^\circ$$, so arc EB $$= 180^\circ - 58^\circ - 70^\circ = 52^\circ$$, i.e. $$\angle EOB = 52^\circ$$. As $$OE = OB$$, $$x = \frac{180 - 52}{2} = 64$$.