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MAT360

Solutions to Homework

1. (Chapter 4,#10)
Prove Proposition 4.7: Hilbert’s Euclidean parallel postulate ⇔ If a line intersects one of two
parallel lines, it must also intersect the other.
Deduce a corollary that transitivity of parallelism is equivalent to Hilbert’s Euclidean parallel
postulate.
Solution: First we show that Hilbert’s parallel postulate implies the line crossingcondition.
Suppose we have two parallel lines l and m, and another line t which intersects l. Let P be the
point where l and t intersect. We must show there is a point Q where m and t intersect. But if
there is no such Q, then lines t and m are parallel. However, by Hilbert’s parallel postulate,
there is at most one line which is parallel to m and contains the given point P. Thus we have
acontradiction.
Now let us show the converse: for any pair of parallel lines l and m, we know that if a line t
crosses one, it must also cross the other. We must establish that Hilbert’s parallel postulate
holds under this assumption. So let l be a given line, and P be a point not on it; we must show
there are not two lines m and n passing through P which are parallel to l. If so, then m l,
andso since n crosses m at P, it must cross l. But this contradicts the assumption that l and n
were parallel.
Now we are to deduce that transitivity of parallelism is equivalent to Hilbert’s parallel postulate. That is,
(l m and m n ⇒ l n) ⇔ Hilbert’s parallel axiom.
But observe that the contrapositive of transitivity of parallelism is precisely the statement we
dealt with before. That is, thecontrapositive is
l n ⇒ l m or m n
or, in words, “if l crosses n, then either l crosses m or n crosses m”. If we use t instead of n
and assuming l m, we have the statement above: “if l crosses t, then t crosses m” (since we
cannot have l crossing m).
2. (Ch. 4, #11) Prove that Hilbert’s parallel postulate is equivalent to the converse of the Alternate
Interior Angles theorem.
Solution:First, we assume the converse to AIA, and establish Hilbert’s parallel postulate.
We have a line l and a point P, and want to demonstrate that there is at most one line parallel
to l containing P.

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n
Construct a perpendicular t to l that contains m
P
C
P, and then let m be the line perpendicular to
D
t. For notational purposes, let A be the point
where t and l intersect, B beanother point l,
and C be a point of m on the other side of t l
A
B
t
from B. (See the figure).
Now suppose that Hilbert’s parallel postulate fails to hold, that is, there is another line n
←
→
which contains P and is parallel to l. Let D be a point on n on the opposite side of AP from
B. By the converse of AIA, since n l and ∠BAP and ∠APD are alternate interior angles,
∠BAP ∼ ∠APD. Butthis means that m = n, by congruence axiom C4.
=

n
m

P

C
D

l

B

A

For the other direction, we assume Hilbert’s
parallel postulate and show that whenever
two parallel lines l and m are cut by a
transversal t, the resulting alternate interior
angles are congruent.

t

←
→
←
→
←
→
Suppose then that we have line l =AB cut by a transversal t =AP, with n =PD beingparallel to
←
→
AB. Suppose also, for contradiction, that the alternate interior angles ∠BAP and ∠APD are
←
→
not congruent. Then, by axiom C4, we can create line PC so that ∠BAP ∼ ∠APC. Applying
=
←←
→→
the Alternate Interior Angle theorem (not the converse!), we know that AB PC. But this
contradicts Hilbert’s postulate, since we have two lines containing P that are parallel to l.
3. (Ch4. #14) Fill in the details of Heron’s proof of the triangle inequality (for any triangle
ABC, we have |AB| + |AC| > |BC|).
Solution: Given ABC, bisect ∠A, and let the bisector meet BC at a point D (which must
exist because of the crossbar theorem).
Observe that ∠ADC is an exterior angle to ABD, so ∠ADC > ∠BAD, but ∠BAD = ∠DAC
(since we bisected the angle at A). Thus, |AC| > |DC| since in...