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Circles $S_1$ and $S_2$ with centres $O_1$ and $O_2$ intersect in points A and B. The circle, which goes through points $O_1$, $O_2$ and $A$, intersects the circle $S_1$ in the point $D$ and also intersects the circle $S_2$ in the point $E$. How can I prove that $CD = CB = CE$? enter image description here

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Use that $ACEO_2$ and $ACDO_1$ are cyclics.

$$\measuredangle CBE=180^{\circ}-\measuredangle EBO_2-\measuredangle ABO_2=180^{\circ}-\measuredangle BAO_2-\measuredangle BEO_2=\measuredangle CEB,$$ which gives $CE=CB$.

By the same way we can get that $CD=CB$ and we are done!

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  • $\begingroup$ Wait, how did you get that the sum of the angles BAO_2 and BEO_2 is equal to angle CEB? $\endgroup$ – idliketodothis Apr 12 '17 at 20:07
  • $\begingroup$ @idliketodothis $180^{\circ}-\measuredangle BAO_2-\measuredangle BEO_2=\measuredangle CEO_2-\measuredangle BEO_2=\measuredangle CEB$ $\endgroup$ – Michael Rozenberg Apr 12 '17 at 20:12
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With C as center and CB as radius if a circle is drawn it should pass through D and E because of cyclic symmetry of 3 circles around the common central concurrency point B. Narrow lens like portion has angles $ \alpha+ \beta, \beta +\gamma, \gamma +\alpha $ around B..

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