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Given a triangle with squares on two sides, the line segments joining the centres of the squares to the midpoint of the third side are equal and perpendicular

Given a triangle with squares on two sides, the line segments joining the centres of the squares to the midpoint of the third side are equal and perpendicular

I am reading Tristan Needham's *Visual Complex Analysis* (2012 reprint, OUP), and in $\S$1.III.3: *Geometry*, the author gives a geometric proof of the following fact, shown in Figure [12b] on page 16:

>![Figure[12a] shows an arbitrary quadrilateral with squares constructed on each of the sides and line segments joining the centres of the opposite squares. Figure [12b] shows an arbitrary triangle with squares constructed on two sides and line segments joining the centres of the squares to the midpoint of the remaining side of the triangle. The centres of the squares are labelled $p$ and $m$, and the midpoint of the side is labelled $m$.](https://math.codidact.com/uploads/g6HNDyFmBWV8wjktevJtXDpG)
>
> <sup>Image taken from *Visual Complex Analysis*, Tristan Needham, page 16.</sup>

> Here squares have been constructed on two sides of an arbitrary triangle, and, as the picture suggests, *the line-segments from their centres to the midpoint $m$ of the remaining side are perpendicular and of equal length*.
>
><sup>*Visual Complex Analysis*, Tristan Needham, page 17.</sup>

To this end, the author shows the following fact. Let $\mathcal{T}_v$ be a translation of the plane by $v$, that is, $\mathcal{T}_v(z) = z + v$ for all $z \in \mathbb{C}$. Let $\mathcal{R}_a^\theta$ be a rotation of the plane about $a$ by an angle $\theta$ in the counter-clockwise direction. One can show that $\mathcal{R}_a^\theta(z) = e^{\iota\theta}z+k$ where $k = a(1-e^{\iota\theta})$. Then we have the following:
> *Let $\mathcal{M} = \mathcal{R}_{a_n}^{\theta_n} \circ \dotsb \circ \mathcal{R}_{a_2}^{\theta_2} \circ \mathcal{R}_{a_1}^{\theta_1}$ be the composition of $n$ rotations, and let $\Theta = \theta_1 + \theta_2 + \dotsb + \theta_n$ be the total amount of rotation. In general, $\mathcal{M} = \mathcal{R}_c^\Theta$ (for some $c$), but if $\Theta$ is a multiple of $2\pi$ then $\mathcal{M} = \mathcal{T}_v$ for some $v$.*
>
><sup>*Visual Complex Analysis*, Tristan Needham, page 19.</sup>

Now, the author asks us to consider $\mathcal{M} = \mathcal{R}_m^\pi \circ \mathcal{R}_p^{(\pi/2)} \circ \mathcal{R}_s^{(\pi/2)}$. Since $(\pi/2) + (\pi/2) + \pi = 2\pi$, the above result says that $\mathcal{M} = \mathcal{T}_v$ for some $v$. To find $v$, it suffices to compute $\mathcal{M}(z)$ for any one point.

Next, the author says, "Clearly, $\mathcal{M}(k) = k$", but this is not so clear to me. I presume that $k = a(1 - e^{\iota\theta})$ for some appropriate $a$ and $\theta$, since the letter $k$ is not for any other purpose in this section. But, it is not clear to me which $a$ and $\theta$ I am supposed to take here since $\mathcal{M}$ is a composition of three rotations.

Nevertheless, I tried just taking the origin to see if it is fixed under $\mathcal{M}$. I get
$$
\begin{gather}
\mathcal{R}_s^{(\pi/2)}(0) = s(1 - \iota),\\\\
\mathcal{R}_p^{(\pi/2)}\bigl(s(1-\iota)\bigr) = \iota s (1 - \iota) + p(1 - \iota) = (s + p) + \iota (s - p),\\\\
\mathcal{R}_m^\pi\bigl( (s + p) + \iota (s - p) \bigr) = -(s + p) - \iota (s - p) + 2m.
\end{gather}
$$
How can I easily see that the last expression is equal to $0$ (presuming my calculations are correct)?