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I heard that in euclidean geometry one has to prove trivial things. I was wondering how one could prove this one?

Let $O$ be a center of a circle $\Gamma$. Suppose that $A,B,C,D$ are points on the circumference of $\Gamma$ such that $AB||CD$. Let $M_1$ and $M_2$ be midpoints of $AB$ and $CD$ respectively. Prove that if $EF$ is a chord of $\Gamma$ such that $M_1,M_2\in EF$ then also $O\in EF$.

This looks so trivial that it seems to hard to find a proof because I'm not sure if I'm skipping something.

guest
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1 Answers1

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The perpendicular bisector of $AB$, i.e. the perpendicular to $AB$ through its midpoint, is the locus of points $P$ such that $PA=PB$. By the definition of circle, $O$ belongs to the perpendicular bisector of $AB$. The same happens for the perpendicular bisector of $CD$, but since $CD\parallel AB$, we have that $E, F$ and $O$ are collinear, due to $\widehat{EOF}=\pi$.

That is almost trivial, I agree. However, it is important, since it gives a useful tool:

If an ellipse is depicted on a plane, how can we find its centre by straightedge and compass?

Jack D'Aurizio
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