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This is problem 1.9.5 from Geometry Revisited by Coxeter and Greitzer: If two lines through one vertex of an equilateral triangle divide the semicircle drawn outward on the opposite side into three equal arcs, these same lines divide the side itself into three equal line segments.

I came up with the following solution for it (click to enlarge image if print too small), but I feel like I have made it more complicated than it needs to be. Wondering if anyone can show how to simplify the answer?

solution to 1.9.5 Geometry Revisited

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See the picture below:

enter image description here

Note that: $$\triangle AB'F \sim \triangle ABJ_1 \quad (1)$$ $$\triangle AFG \sim \triangle AJ_1K_1 \quad (2)$$ $$\triangle AGC' \sim \triangle AK_1C \quad (3)$$ As triangles $\triangle ABJ_1$, $\triangle AJ_1K_1$, $\triangle AK_1C$ share common sides and triangles $\triangle AB'F$, $\triangle AFG$ and $\triangle AGC'$ too, the similarity ratio $k$ is the same for $(1)$, $(2)$ and $(3)$. In this way we have: $$BJ_1=k\,B'F = k \, r$$ $$J_1K_1=k\,FG = k \, r$$ $$K_1C=k\,GC = k \, r$$

Therefore

$$BJ_1=J_1K_1=K_1C.$$

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  • $\begingroup$ Thanks, that is a much simpler solution. $\endgroup$ – Circulwyrd Jun 4 '13 at 1:53

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