a line segment has at most one midpoint

(this proof is not correct yet)

Theorem 1.

In an ordered geometry a line segment has at most one midpoint.

Proof.

Let $[p,q]$ be a closed line segment and suppose $m$ and $m^{\\prime}$ are midpoints.If $m : p : q$ then $[m,p]<[m,q]$ so $m$ is not a midpoint. Similarly we cannot have $p : q : m$ , so we have $p : m : q$ . And also, $p:m^{\\prime}:q$ . Suppose $m\ot=m^{\\prime}$ . Without loss ofgenerality we can assume $p:m:m^{\\prime}$ and $m:m^{\\prime}:q$ . But then $[p,m^{\\prime}]>[p,m]\\cong[m,q]>[m^{\\prime},q]$ so that $[p,m^{\\prime}]\ot\\cong[m^{\\prime},q]$ , a contradiction. Hence $m=m^{\\prime}$ .∎

单词	ALineSegmentHasAtMostOneMidpoint
释义	a line segment has at most one midpoint (this proof is not correct yet) Theorem 1. In an ordered geometry a line segment has at most one midpoint. Proof. Let $[p,q]$ be a closed line segment and suppose $m$ and $m^{\\prime}$ are midpoints.If $m : p : q$ then $[m,p]<[m,q]$ so $m$ is not a midpoint. Similarly we cannot have $p : q : m$ , so we have $p : m : q$ . And also, $p:m^{\\prime}:q$ . Suppose $m\ot=m^{\\prime}$ . Without loss ofgenerality we can assume $p:m:m^{\\prime}$ and $m:m^{\\prime}:q$ . But then $[p,m^{\\prime}]>[p,m]\\cong[m,q]>[m^{\\prime},q]$ so that $[p,m^{\\prime}]\ot\\cong[m^{\\prime},q]$ , a contradiction. Hence $m=m^{\\prime}$ .∎
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