### Proof of Lemma 1

**Lemma 1**

Suppose that **t** is an isometry which fixes the points O(0,0) and X(1,0). Then either
**t
** is the identity, or
**t** = **r**_{x} (reflection in the x-axis).

**Proof**

Let P be the point (x,y), and suppose that **t**(P) = Q, the point (u,v).
As **t** fixes O, and is an isometry, |OP| = |OQ|.

Squaring, we get x^{2} + y^{2} = u^{2} + v^{2}.

Similarly, |XP| = |XQ|, so (x-1)^{2} + y^{2} = (u-1)^{2} + v^{2}.

Combining these, we get -2x + 1 = -2u + 1, so that u = x.

Let Y be the point (0,1), and let Y' = **t**(Y). Since **t** is an isometry, |OY'| = 1.

As OY is perpendicular to OX and **t** fixes OX, OY' is perpendicular to OX.

Thus Y' is on the y-axis.

As |OY'| = 1, Y' = Y or Y_{-} = (0,-1).

In the first case, we have |PY| = |QY|, so, as above, x^{2} + (y-1)^{2} = u^{2} + (v-1)^{2},

and we deduce that v = y.

Thus, in this case (u,v = (x,y), so **t** is the identity.

In the second case, we have |PY| = |QY_{-}|, and we deduce that v = -y.

Now we have (u,v) = (x,-y), the reflection of (x,y) in the x-axis.