what is the angle here in this cyclic triangle?

If you draw a line from the center of the circle to your angle, you get 2 isoscele triangles, so your angle is the sum of those 2 other angles in this quadrangle. the sum of all angles of quadrangle is $$360^{\circ}$$ so your angle is $$\frac{360^{\circ}-90^{\circ}}{2} = 135^{\circ}$$

Explanation:

$$AD = AB = AC$$ $$\angle{ADB} = \angle{ABD} ; \angle{ADC} = \angle{ACD}$$ $$\angle{BDC} = \angle{ADB} + \angle{ADC} = \angle{ABD} + \angle{ACD}$$ $$\angle{BAC} + \angle{ABD} + \angle{BDC} + \angle{ACD} = 360^{\circ}$$ $$90^{\circ} + 2\angle{BDC} = 360^{\circ}$$ $$\angle{BDC} = \frac{360^{\circ}-90^{\circ}}{2} = 135^{\circ}$$

Draw the rest of circle (complete the circle),

Inscribed Angle =(1/2)Intercepted Arc

$$m(\widehat{CDB})=\frac{1}{2}m( \stackrel \frown{CEB})$$ $$m(\widehat{CDB})=\frac{1}{2}270^{o}=135^{o}$$