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Given two rays, L and M, with common origin O, and a point Q inside the acute angle formed by the rays, reflect Q across L to obtain Q' and then reflect Q' across M to obtain Q''. Similarly, reflect Q across M to obtain P' and then reflect P' across L to obtain P''. Show that the line through O and Q is the perpendicular bisector of the segment joining P'' and Q''. How do you do this? It is easy to show that Q and all reflected points lie on the circle with center O that passes through Q. I thought of using reflection across the line through O and Q and then use a symmetry argument but couldn't see just how to implement this idea. I would really appreciate any help. Thanks

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It would help some of us if you could provide a picture. – NoChance Sep 18 '12 at 22:42
I'm sorry I didn't provide a picture but I don't know how to do this using a programming language and as a new user I am not allowed to post images – samuel Sep 18 '12 at 22:54
@Emmad: I’ve taken the liberty of adding a diagram. – Brian M. Scott Sep 18 '12 at 23:42
@Thanks Brian, this is helpful. – NoChance Sep 18 '12 at 23:55
If P" and Q" lie on the premituer of a circle with O at the center, then if you draw 2 lines, OP" and OQ" these 2 lines would be equal (each is a radius of the same circle). From this the proof follows. – NoChance Sep 19 '12 at 1:01

Thanks Emmad. As you said, the result follows. The fact you mention combined with the following show that the triangles that need to be shown are congruent in fact are: the angle Q''Q'Q equals the angle P''P'Q so the arcs they intercept are the same. Thank you for your answer

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This should be a comment, not an answer – Hagen von Eitzen Oct 16 '12 at 22:10

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