regular 7-gons problem There are two regular 7-gons sharing one vertex. The other vertices of the first one (starting from the common vertex $A_0(B_0)$)  in (say) clockwise order are $A_1, A_2... A_6$. The other vertices of the second one in the same order (clockwise in our case) are $B_1...B_6$. Show that the lines $A_1B_1,A_2B_2...A_6B_6$ go through one point. There is a hint: What similarity transformation takes one 7-gon to the other?

 A: As pointed out by Narasimhan the common point of intersection of the seven lines connecting corresponding vertices of the two 7-gons is the other point of intersection of the circles, call them $\color{orange}C_1$ and $\color{blue}C_2$, circumscribing the 7-gons. This is not using the hint, so possibly useless to the OP.
That claim follows relatively easily from basic circle geometry.
Let $O_1$ and $O_2$ be the centers of $C_1$ and $C_2$ respectively. Let $A$ and $P$ be the points where the two circles intersect, so $A$ is a common vertex of the two polygons.

Let $L$ be any line through $P$. Denote by $Q$ (resp. $R$) the intersection of $L$ and $C_1$ (resp. $L\cap C_2$). Let $\alpha$ be the central angle $\angle AO_1Q$ (think of this in the interval $[0,2\pi]$ and specify that $AO_1$ is the left ray of that angle and $QO_1$ the right ray.
Then the inscribed angle (of $C_1$) equals $\angle APQ=\alpha/2$. We then get the same angle between the ray $PR$ and the extension (beyond $P$) of the line segment $AP$. Therefore the inscribed angle (in $C_2$ now!) $\angle RPA=\pi-\alpha/2$. This means that the central angle 
$$
\angle RO_2A=2(\pi-\frac\alpha2)=2\pi-\alpha
$$
(here $A$ lies on the right ray, and $R$ on the left ray).
Consequently the central angle $\angle AO_2R=\alpha$ (with left and right reversed) as well.
Therefore we have proven:

With the above notations and definitions (cf. figure above) we have equality of the central angles $\angle AO_1Q=\angle AO_2R$.

The OP's claim follows from this as follows. We let $Q$ run over the set of vertices of the $7$-gon circumscribed by $C_1$. This amounts to letting $\alpha=k (2\pi/7)$ for $k=1,2,3,4,5,6.$ The above result then tells us that the corresponding sequence of points $R$ are the vertices of the unique regular $7$-gon circumscribed by $C_2$ that has $A$ as one of its vertices.
