If I join the chord then I am getting the angles of the triangle are $45,45,90$ so $\theta=180-2x$ where $x$ is the angle of the other triangle whose angle is $\theta$
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4$\begingroup$ Hint: Inscribed Angle Theorem. (It may help to draw the full circle.) $\endgroup$ – Blue Sep 26 '13 at 6:45
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$\begingroup$ The central angle is twice the inscribed angle. en.wikipedia.org/wiki/Inscribed_angle#Theorem $\endgroup$ – Sammy Black Sep 26 '13 at 6:46
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
$$m(\widehat{CDB})=\frac{1}{2}m( \stackrel \frown{CEB})$$
$$m(\widehat{CDB})=\frac{1}{2}270^{o}=135^{o}$$