# what is the angle here in this cyclic triangle?

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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Hint: Inscribed Angle Theorem. (It may help to draw the full circle.) – Blue Sep 26 '13 at 6:45
The central angle is twice the inscribed angle. en.wikipedia.org/wiki/Inscribed_angle#Theorem – 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:

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

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Draw the rest of circle (complete the circle),

Inscribed Angle =(1/2)Intercepted Arc


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

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