Problem 1539

1. **Identify the common point $M$:**
Let $M$ be the common point of the circumcircles of $\triangle ABC$ and $\triangle ADE$. Since $M$ lies on both circumcircles, it must be the center of the spiral similarity that maps $BD$ to $CE$. Given that $BD = CE$, $M$ must be the midpoint of the arc $BAC$ on the circumcircle of $\triangle ABC$ that does not contain $A$.

2. **Angle chasing to show $\angle O_1OA = \angle O_2OM$:**
Since $M$ is the midpoint of the arc $BAC$, we have:
$$\angle BAM = \angle CAM$$
This implies that:
$$\angle BAM = \angle CAM = \frac{1}{2} \angle BAC$$
Since $O_1$ and $O_2$ are the circumcenters of $\triangle ABE$ and $\triangle ACD$ respectively, we have:
$$\angle O_1AO = \angle O_2AO = 90^\circ - \frac{1}{2} \angle BAC$$
Therefore:
$$\angle O_1OA = \angle O_2OM$$

3. **Prove that $OO_1 = OO_2$:**
To show that $OO_1 = OO_2$, we observe the following:
- $\triangle BOO_1 \sim \triangle BEC$
- $\triangle COO_2 \sim \triangle BED$

Using the Law of Sines in $\triangle ABE$ and $\triangle ACD$, we get:
$$\frac{OO_1}{OO_2} = \frac{|DC| \cdot |BO_1|}{|BE| \cdot |CO_2|}$$
Since $BD = CE$, it follows that:
$$\frac{|DC|}{|BE|} = 1$$
Therefore:
$$\frac{OO_1}{OO_2} = 1 \implies OO_1 = OO_2$$

4. **Conclude that $AO_1O_2M$ is cyclic:**
Since $OO_1 = OO_2$ and $\angle O_1OA = \angle O_2OM$, we have:
$$\triangle AOO_1 \sim \triangle MOO_2$$
and
$$\triangle MOO_1 \sim \triangle AOO_2$$
This implies that $AO_1O_2M$ is an isosceles trapezoid, which is cyclic.

Therefore, the circumcircles of $\triangle ABC$, $\triangle ADE$, and $\triangle AO_1O_2$ have a common point different from $A$.

$\blacksquare$