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Prove the composite of three rotations of 180° is a rotation of 180°. ( In $\Bbb E^2$, the rotation is about 3 different points).

I don't really know where to get started. I have proven that every composite of two rotations of 180° is a translation. So this would be translation+a rotation of 180° or a rotation of 180° + translation. I can see why this is true but I'm stuck at proving it.

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    In two dimensional space rotation by $540^\circ$ is simply the same as by $540^\circ-360^\circ=180^\circ$ . Translation doesn't affect the rotation, it affects only the final translation. – 2017-02-23
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    I see, but I have to prove it, and I don't think that's sufficient for a proof – 2017-02-23
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    We can prove it starting from homogeneous matrix composing rotation and translation. But I don't know whether you are familiar with this notation.. – 2017-02-23
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    I've seen them a couple of times. I'll try it! – 2017-02-23
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    I can't understand why translations are involved in a problem concerned only with rotations... – 2017-02-23

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@Widawensen

It has not been remarked that a $180^{\circ}$ rotation around a center $C$ is plainly a central symmetry with respect to this point. This allows a simple proof using vectors only ; moreover, this proof provides the final center of symmetry $C_4$ (see figure below).

The adjective "central" is there to differentiate these transformations in particular from orthogonal symmetries with respect to a line.

Denoting by $S_1, S_2, S_3$ the successive central symmetries, by $C_1,C_2,C_3$ their respective centers, by $M$ a generic point in the plane, by $M'=S_1(M), M''=S_2(M'), M'''=S_3(M'')$ its successive images, we have:

$$\begin{cases}\vec{C_1M'}&=&-\vec{C_1M}\\\vec{C_2M''}&=&-\vec{C_2M'}\\\vec{C_3M'''}&=&-\vec{C_3M''}\end{cases} \ \iff \ \begin{cases}M'-C_1&=&C_1-M\\M''-C_2&=&C_2-M'\\M'''-C_3&=&C_3-M''\end{cases}$$

$$\iff \ \begin{cases}M'&=&2C_1-M\\M''&=&2C_2-M'\\M'''&=&2C_3-M''\end{cases}$$

Thus, we have:

$$M'''=2C_3-(2C_2-(2C_1-M))) \ \iff \ M'''=2C_3-2C_2+2C_1-M \ \iff$$

$$\dfrac{M+M'''}{2}=\underbrace{C_3-C_2+C_1}_{C_4}$$

Thus, $C_4$ defined by

$$\tag{1}C_4:=C_3-C_2+C_1 \ \ \iff \ \ \vec{C_1C_4}=\vec{C_2C_3}$$

is the midpoint of $M$ and $M'''$ (see picture).

In other words, $M'''$ is the symmetrical point of $M$ with respect to $C_4.$

As a conclusion, the composition $S_3\circ S_2\circ S_1$ of three central symmetries with respect to points $C_1,C_2,C_3$, is a central symmetry with respect to point $C_4$ defined by (1).

Remark: there is a nice 3D interpretation of this figure ; imagine it as a perspective view of a "contemporary style" bench with a rectangular sitting zone $C_1C_2C_3C_4$, and bench's feet in $M$ and $M''$.

enter image description here

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    I have added an image and a 3D geometrical interpretation. – 2017-02-23
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    So this rotation as the symmetry leads really to more geometrical interpretation ... very interesting construction.. – 2017-02-24
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    And here also: $C_4= C_1+R_1{C_2}+R_1{R_2}C_3$ – 2017-02-24