Download Homework 2 solutions 2.6, #3 Outline how SMSG Postulate 4 can be

Homework 2 solutions
2.6, #3 Outline how SMSG Postulate 4 can be deduced from the other SMSG
postulates.
Outline: Let P and Q be points on a line. The ruler postulate guarantees that the
point of the line can be placed in 1-1 correspondence with the real numbers. Let
x P and x Q be the real numbers associated with P and Q under this
correspondence. Define a 2nd correspondence of the points on the line with the
real numbers by setting the real number associated with a point A, y A , to be
x A  x P for each point A on the line. Then this second correspondence assigns the
number 0 to P. Finally, if the coordinate assigned to Q, y Q , is positive, we are
done. Otherwise, replace each coordinate y A by its negative to obtain a third
correspondence that satisfies Postulate 4.
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2.6, # 5 Which of Birkhoff’s Axioms implies SMSG Postulate 15?
Postulate IV with k  1.
3.2, # 8 An equilateral triangle has three congruent sides. Prove that an equilateral
triangle is equiangular.
Proof: Let ABC be equilateral. Consider angles A and B. By Theorem 3.2.7,
these angles are congruent since the sides opposite these angles are congruent.
Similarly, angles A and C are congruent. So all three angles are congruent.

3.2, # 9 Prove that if a point P is on the perpendicular bisector of a line segment AB,
then it is equidistant from the endpoints of the line segment.
Proof: Let P be on the perpendicular bisector of a line segment AB. Let M be the
midpoint of the segment. Consider the triangles AMP and BMP. Side AM is
congruent to side BM since M is the midpoint. Angles AMP and BMP are both
right angles, so they are congruent. Side MP of the first triangle is congruent to
side MP of the second. So by SAS, the triangles are congruent. So sides AP and
BP are congruent. Thus P is equidistant from A and B.
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