# About the Three Reflections Theorem

I recently solved this exercise from Hartshorne's Classical Geometry.

Given three lines $a$, $b$, $c$ through a point $O$, show that there exists a unique fourth line $d$ such that $$\sigma_c\sigma_b\sigma_a=\sigma_d,$$ where $\sigma$ denotes the reflection in a given line.

In the diagram below, I take $A\in a$, let $B$ be the reflection of $A$ across $b$, $C$ be the reflection of $B$ across $c$, and then let $d$ be the perpendicular bisector of $AC$. I was able to prove $\sigma_c\sigma_b\sigma_a=\sigma_d$, which I will be happy to include if someone asks. Anyway, based on that, I know that $\sigma_d$ fixes $O$, and thus passes through $O$.

My follow up question to this exercise is, is the angle between $a$ and $d$ congruent to the angle between $b$ and $c$? I'm curious to know, because if so, I believe this implies that given three reflection axes through a point, we can rotate two of them so that one coincides with the third, hence cancelling, and the image of the first rotated line will be the desired $d$.

I have a hunch it is true based on an earlier question, but this seems like a different way of proving it. I've included a picture with which angles I know are congruent, and I assume I'm working in a Euclidean plane. Thanks for any insight.

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as $\sigma_c\sigma_b=\sigma_d\sigma_a$, the angles must be the same (composition of 2 reflections is the rotation by twice the angle) –  user8268 Apr 7 '11 at 7:10
@user8268, ah, so simple! I think I was trying too hard to appeal to cyclic quadrilaterals and supplementary arguments in my head. I'd be happy to accept if you post as an answer. –  yunone Apr 7 '11 at 7:13
@user8268: Please write answers as answers and not as comments (even if they are simple). –  Hans Lundmark Apr 7 '11 at 7:20
ah, sorry ... I'll post it as answer –  user8268 Apr 7 '11 at 8:02
@user8268: I used to write simple answers as comments, until @Hans convinced me otherwise by pointing me to this thread: meta.math.stackexchange.com/questions/1559/… –  joriki Apr 7 '11 at 8:17

as $\sigma_c\sigma_b=\sigma_d\sigma_a$, the angles must be the same (composition of 2 reflections is the rotation by twice the angle)

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The answer is yes. I don't have a purely geometric proof, since I cheated and used linear algebra. :)

Let $T_\phi = \begin{pmatrix} \cos\phi & \sin\phi \\ \sin\phi & -\cos\phi \end{pmatrix}$. This matrix represents reflection across a line at an angle of $\phi/2$ to the $x$ axis.

If your lines $a$, $b$, $c$ have angles $\alpha/2$, $\beta/2$, $\gamma/2$ to the $x$ axis, then $\sigma_c \sigma_b \sigma_a$ is represented by the matrix $T_\gamma T_\beta T_\alpha$, which equals $T_{\gamma-\beta+\alpha}$ by pure algebra. We can choose the coordinate system so that $a$ is the $x$ axis, and hence $\alpha=0$. Then, by the matrix formula above, $d$ lies at an angle of $(\gamma-\beta+0)/2$ to the $x$ axis (that is, to $a$), and this is the same as the angle $\gamma/2-\beta/2$ between $c$ and $b$.

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Argh, again! Someone posted a perfect answer as a comment (which other answerers don't get any notification about), and the OP says "please post this as an answer so that I can accept it". See the discussion from yesterday here: math.stackexchange.com/q/31324/1242 –  Hans Lundmark Apr 7 '11 at 7:19
I upvoted it anyway. :D I was going to use coordinate geometry myself, but I do not have the time to write out the equations... –  Guess who it is. Apr 7 '11 at 7:22
+1, thanks for your answer Hans. I was hoping for a purely geometric argument, because I'm not assuming any underlying ordered field on the geometry, so I may not be able to impose a coordinate system (I think). I appreciate this way though in the case that I do have a coordinate system. –  yunone Apr 7 '11 at 7:58