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vantroff
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Hi, the question is from Serge Lang - Basic mathematics, Page 171 exercise 6.
Thing to prove:
My first attempt by using the just the Pythagorean theorem.
Second attempt by using isometries (not using Theorem 10 from the book).
Can the red part be replaced by this blue part (or omitted)?
Angle ∠MQP=90° and ∠P'QM=90° ⇒ ∠ P'QP is 180° ⇒ P,Q,P' are collinear.
The second proof is more important to me because I'm at the Isometries part of the book. Is there something wrong with these proofs?
Thanks in advance for everyone who will spend time on this.
Thing to prove:
Let ΔPQM and ΔP'Q'M' be right triangles whose right angles are at Q and Q', respectively. Assume that the corresponding legs have the same length:
d(P,Q)=d(P',Q')
d(Q,M)=d(Q',M')
Then the right triangles are congruent.
d(P,Q)=d(P',Q')
d(Q,M)=d(Q',M')
Then the right triangles are congruent.
My first attempt by using the just the Pythagorean theorem.
d(Q,M)2 + d(P,Q)2 = d(P,M)2
d(Q,M)2 + d(P,Q)2 = d(Q',M')2 + d(P',Q')2 = d(P',M')2
It follows that:
d(P,M)2 = d(P',M')2
and
d(P,M) = d(P',M')
We conclude that the triangles are congruent by Theorem 10 in the book (page 168) which states the following:
Let ΔP,Q,M and ΔP',Q',M' be triangles whose corresponding sides have the equal length, that is
d(P,M)=d(P',M')
d(P,Q)=d(P',Q')
d(Q,M)=d(Q',M')
These triangles are congruent.
d(Q,M)2 + d(P,Q)2 = d(Q',M')2 + d(P',Q')2 = d(P',M')2
It follows that:
d(P,M)2 = d(P',M')2
and
d(P,M) = d(P',M')
We conclude that the triangles are congruent by Theorem 10 in the book (page 168) which states the following:
Let ΔP,Q,M and ΔP',Q',M' be triangles whose corresponding sides have the equal length, that is
d(P,M)=d(P',M')
d(P,Q)=d(P',Q')
d(Q,M)=d(Q',M')
These triangles are congruent.
Second attempt by using isometries (not using Theorem 10 from the book).
There exist translation which maps Q on Q'. Hence it suffices to prove our assertion when Q=Q'. We now assume Q=Q'. Since d(Q,M)=d(Q,M') there exist rotation relative to Q which maps M on M'. This rotation leaves Q fixed.
This reduces the case to :
Q=Q'
M=M'
Now either P=P' or P≠P'. We assume P≠P'.
Let LQM be the line passing through point Q and M.
The segments P'Q and PQ are perpendicular to LQM. Also, d(Q,P')=d(Q,P).
⇒ Q lies on the perpendicular bisector of the segment PP'. The only perpendicular line to PP' which passes through Q is LQM. We conclude that LQM is the perpendicular bisector of PP' and that poins P,Q,P' are collinear and P is at the same distance from Q as P' but in the opposite direction.
⇒There exist reflection through LQM which leaves LQM fixed and maps P on P'.
We found that there exist composite isometry F such that,
F(Q)=Q'
F(M)=M'
F(P)=P'
and given that the image of line segment PQ under isometry G is the line segment between G(P) and G(Q), we conclude the assertion proven.
This reduces the case to :
Q=Q'
M=M'
Now either P=P' or P≠P'. We assume P≠P'.
Let LQM be the line passing through point Q and M.
The segments P'Q and PQ are perpendicular to LQM. Also, d(Q,P')=d(Q,P).
⇒ Q lies on the perpendicular bisector of the segment PP'. The only perpendicular line to PP' which passes through Q is LQM. We conclude that LQM is the perpendicular bisector of PP' and that poins P,Q,P' are collinear and P is at the same distance from Q as P' but in the opposite direction.
⇒There exist reflection through LQM which leaves LQM fixed and maps P on P'.
We found that there exist composite isometry F such that,
F(Q)=Q'
F(M)=M'
F(P)=P'
and given that the image of line segment PQ under isometry G is the line segment between G(P) and G(Q), we conclude the assertion proven.
Can the red part be replaced by this blue part (or omitted)?
Angle ∠MQP=90° and ∠P'QM=90° ⇒ ∠ P'QP is 180° ⇒ P,Q,P' are collinear.
The second proof is more important to me because I'm at the Isometries part of the book. Is there something wrong with these proofs?
Thanks in advance for everyone who will spend time on this.
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