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Proof: Consider the segment PQ. We extend it to the line PQ. Thus we can apply the ruler axiom, with P,Q corresponding to real numbers x.y. Then there is exactly one number equidistant to X.ỵ, namely (x+y)/2. Moreover, there is exactly one point corresponding to (x+y)/2, say M. We claim that this is the unique midpoint of PQ. By definition, M is between P and Q, and d(P,M)=(x±y)/2-x=(y-x)/2=y-(x±y)/2=d(M,Q). Another way to prove the uniqueness of the midpoint is to assume that there are two midpoints M and N, then conclude that M=N. Prove the midpoint theorem using this method.

Proof: Consider the segment PQ. We extend it to the line PQ. Thus we can apply the ruler axiom, with P,Q corresponding to real numbers x.y. Then there is exactly one number equidistant to X.ỵ, namely (x+y)/2. Moreover, there is exactly one point corresponding to (x+y)/2, say M. We claim that this is the unique midpoint of PQ. By definition, M is between P and Q, and d(P,M)=(x±y)/2-x=(y-x)/2=y-(x±y)/2=d(M,Q). Another way to prove the uniqueness of the midpoint is to assume that there are two midpoints M and N, then conclude that M=N. Prove the midpoint theorem using this method.

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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Proof: Consider the segment PQ. We extend it to the line PQ. Thus we can apply the ruler axiom, with P,Q corresponding to real numbers x.y. Then there is exactly one number equidistant to XV. namely (x+Y)/2. Moreover, there is exactly one point corresponding to (X+y)/2, say M. We claim that this is the unique midpoint of PQ. By definition, M is between P and Q, and d(P,M)=(x±y)/2-x=(y-x)/2=y-(x±y)/2=d(M,Q). Another way to prove the uniqueness of the midpoint is to assume that there are two midpoints M and N, then conclude that M=N. Prove the midpoint theorem using this method.
Transcribed Image Text:Proof: Consider the segment PQ. We extend it to the line PQ. Thus we can apply the ruler axiom, with P,Q corresponding to real numbers x.y. Then there is exactly one number equidistant to XV. namely (x+Y)/2. Moreover, there is exactly one point corresponding to (X+y)/2, say M. We claim that this is the unique midpoint of PQ. By definition, M is between P and Q, and d(P,M)=(x±y)/2-x=(y-x)/2=y-(x±y)/2=d(M,Q). Another way to prove the uniqueness of the midpoint is to assume that there are two midpoints M and N, then conclude that M=N. Prove the midpoint theorem using this method.
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