Reim's Theorem

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Joined: Jan 12, 2011

by iarnab_kundu, Nov 29, 2013, 6:04 PM

The following is a repost of the result. I found this one in IDMasterz's blog and found it worth mentioning once again. It appears to be obvious at first sight but its a profound theorem.

IDMasterz wrote:

Theorem -

Circles $\omega_1, \omega_2$ intersect at $A, B$ . Line through $A, B$ intersect $\omega_1, \omega_2$ at $C, E$ and $D, F$ respectively. Prove $CE \parallel DF$ .

Proof -
Suppose that the lines intersect at $P$ (can be at infinity). Then $AB$ is anti parallel to $CD$ and $AB$ is also anti parallel to $EF$ hence $EF \parallel CD$ .

Converse -

Suppose we have two circles $\omega_1, \omega$ _2 that intersect at two point $A,B$ . Let the line through $B$ intersect the two circles at $C, D$ ( $\omega_1, \omega_2$ respectively). Then suppose a line that passes cuts $\omega_1$ at $E$ and $\omega_2$ at $F$ in a way that $CE \parallel DF$ . Then that line passes through $A$ .

Proof - Suppose we extend the two lines to meet at $X$ . Let the second intersection of the latter line with $\omega_1$ be $A'$ . Then $A'B$ is anti parallel to $CE$ , so $DF$ is anti-parallel to $A'B$ by parallelism, giving $A'$ lies on $\omega_2$ so $A \equiv A'$ .