MAT 3271: Selected solutions to problem set 9
Chapter 3, Exercises:
32. The difficulty in proving the SSS criterion is that the only congruence criteria we have at this
point are SAS and ASA, which both, obviously, require us to know at least one pair of congruent
angles. But we are given no information about angles! What to do? Clearly, we must prove that a
pair of angles is congruent by using what we know about the corresponding sides of the triangles.
The only theorem we have which proves that angles are congruent given that sides are congruent
is Proposition 3.10, the “base angles theorem.” Therefore, our strategy has to be to set up a figure
with some isosceles triangles in it. Now for the details.
Given triangles 4ABC and 4DEF with AB 'DE,B C 'EF , and AC 'DF , let Gbe the
unique point on the opposite side of ←−→
AC from Bsuch that 4DEF ' 4AGC (Corollary to SAS).
Since congruence is transitive for segments and angles, it immediately follows from the definition
that it is transitive for triangles; therefore, it suffices to prove 4ABC ' 4AGC. Furthermore, by
the definition of congruence for triangles and CA2 it follows, of course, that AB 'AG,AC 'AC,
and BC 'GC . We may thus assume without loss of generality that A=D,G=E, and C=F.
That is what the text means by “reduce to the case that A=D,C=F, and the points Band
Eare on opposite sides of ←−→
AC .”
By definition of opposite sides, BE intersects ←−→
AC at a point G(not to be confused with our
earlier use of a point G, now discarded). By BA3, either G=A,G=C,A∗G∗C,G∗A∗C,
or A∗C∗G. The same proof works for the first two cases by exchanging the labels of Aand C.
Similarly, the same proof applies to the last two cases. Thus it suffices to consider three cases.
Case 1. A∗G∗C. We have ∠EBC '∠B EC and ∠E BA '∠B EA by Proposition 3.10. Ray
−−→
BE is between rays −−→
BA and −−→
BC by Proposition 3.7. Similarly, ray −−→
EB is between rays −−→
EA
and −−→
EC . So ∠AB C 'AEC by Proposition 3.19 (Angle Addition). By SAS, 4AB C ' 4AEC .
Case 2. G=A. Just apply Prop. 3.10 and SAS. (Note that it follows from the definition of
congruence for triangles, the definition of a right angle, and the definition of perpendicular that
BE ⊥AC , a useful fact about the median from the apex of an isosceles triangle.)
Case 3. Similar to Case 1, but use angle subtraction instead of angle addition.
Chapter 4, Exercises:
4. Let land mbe parallel lines, and assume line nintersects line mat point P. If ndoes not intersect
l, then nand mare distinct lines through Pthat are parallel to l, contradicting Hilbert’s Parallel
Postulate. Thus Hilbert’s Parallel Postulate implies that if a line intersects one of two parallel
lines, it intersects the other. Conversely, let lbe a line and Pa point not on l. Suppose there are
two lines through P,mand n, that are parallel to l. Then nintersects one of two parallel lines
(m) but not the other (l).
5. Assume Hilbert’s Parallel Postulate, and let land mbe parallel lines cut by a transversal tat
points Pand Q, respectively. Let Rand Sbe p oints on land m, respectively, that are on opposite
sides of t, so ∠RP Q and ∠SQP are alternate interior angles (IA2, BA2). By CA4, there is a
unique ray −−→
P T on the same side of tas Rsuch that ∠TPQ'∠S QP . By the Alternate Interior
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