# Prove that three line segments are equal by using the circles

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 $CD = CB = CE$?

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!
• @idliketodothis $180^{\circ}-\measuredangle BAO_2-\measuredangle BEO_2=\measuredangle CEO_2-\measuredangle BEO_2=\measuredangle CEB$ Apr 12, 2017 at 20:12
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..