Let $K$ be the intersection point of $B\Delta$, $\Gamma E$ and $N$ be the second intersection point of $c_1$, $c_2$. Let $Z$ be the point of intersection of the tangents at $B, \Gamma$ of the circle $c$. We will prove that the points $K, N, Z$ are collinear.
Let $T$ be the intersection of $BZ$ with $c_1$ and $P$ the intersection of $Z\Gamma$ with $c_2$, then:
$$\hat{E}_1 = \hat{T}_1 : (\text{inscribed in } c_1 \text{ at the same arc } BK)$$
$$\hat{E}_1 = \hat{B}_1 : (\text{chord and tangent at } c)$$
$$\hat{\Delta}_1 = \hat{P}_1 : (\text{inscribed in } c_2 \text{ at the same arc } \Gamma K)$$
$$\hat{\Delta}_1 = \hat{\Gamma}_1 : (\text{chord and tangent at } c)$$
$$\hat{B}_1 = \hat{\Gamma}_1 : (\text{ZB and Z\Gamma are tangents at } c)$$
From the above inequalities we have:
$$\hat{B}_1 = \hat{T}_1, \text{ so } KT // B\Gamma \text{ and } \hat{P}_1 = \hat{\Gamma}_1, \text{ so } KP // B\Gamma.$$
Therefore $B\Gamma PT$ is an isosceles trapezium and thus it is cyclic (let $c_3$ its circle).
This means that the common chord $KN$ of the circles $c_1$ and $c_2$ will pass through the radical center $Z$ of the circles $c_1, c_2, c_3$.
Suppose first that $A, K, N$ are collinear. Then, since $K, N, T$ are collinear, we have that $A, K, N, T$ are collinear, so $A, K, N$ are on the symmedian $AT$.
Conversely, if $K$ is on the symmedian $AT$, then $A, K, T$ are collinear and since $K, N, T$ are collinear, we conclude that $A, K, N$ are collinear.

Let $K$ be the intersection point of $B\Delta$, $\Gamma E$ and $N$ the second common point of $c_1, c_2$. Let $Z$ be the point of intersection of the tangents at $B, \Gamma$ of the circle $c$. If $H$ is the intersection of $EB$ and $\Delta\Gamma$, we will prove that $K, N, Z, H$ are collinear.
Using Pascal's theorem at the degenerate hexagon $BB\Gamma\Gamma\Delta E$, we have that $H, K, Z$ are collinear.
The point $H$ has power with respect to $c$, $HE \cdot HB = H\Delta \cdot H\Gamma$.
Therefore $K, N, H$ are collinear. Therefore $K, N, Z, H$ are collinear.

Now it is clear that $A, K, N$ are collinear if and only if $K$ is on the symmedian $AZ$.