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enter image description here Two circles $C_1$ and $C_2$ in the plane intersect at two distinct points $A$ and $B$ , and the centre of $C_2$ lies on $C_1$. Let points $C$ and $D$ be on $C_1$ and $C_2$, respectively, such that $C$,$B$, and $D$ are collinear. Let point $E$ on $C_2$ be such that $DE$ is $parallel$ to $AC$. How do I show that $\vert AE\vert = \vert AB\vert$.

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  • $\begingroup$ What have you tried so far? The first thing I would always do in a geometry question like this is draw a figure. Have you drawn a figure? If you, can you snap it & attach as an image to the question? $\endgroup$ – Colm Bhandal Dec 10 '15 at 16:41
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HINT:

$\angle BAE = \angle BDE$ ( same arc) and $\angle ACB = \angle BDE$ (parallels), so $\angle ACB = \angle BAE =\frac{1}{2} m(AB)$ . However, $\angle BAO_2 = \frac{1}{4} m(AB)$. Hence, $\angle BAE = 2 \angle BAO_2$, and so $E$ is the symmetrical of $B$ wr to $AO_2$ ...

Note $E$ does not vary when $C$ moves on $C_1$.

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