MAT 3271: Selected solutions to problem set 6
Chapter 3, Exercises:
3. (b) We are to prove that AC ⊆ AB ∪ BC. Thus, let P ∈ AC. By definition of union, we must
prove that P ∈ AB or P ∈ BC, which we will do by showing that P ∈
/ AB ⇒ P ∈ BC.
←−→
Let D be a point not lying on AB (Proposition 2.3). By Proposition 2.1, the only point
←−→
←−→
lying on both AB and P D is P . So if P ∈
/ AB, then A and B are on the same side of
←−→
←−→
P D . Since P ∈ AC (given), A and C are on opposite sides of P D . Therefore, B and C
←−→
are on opposite sides of P D by the corollary to BAIV. By definition of opposite sides (and
the observation above regarding Proposition 2.1), P ∈ BC.
4. (a) We prove the contrapositive: Given A∗B ∗C, A∗C ∗P ⇒ A∗B ∗P . This follows immediately
from Proposition 3.3.
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−−−→
(b) The definition of ray is equivalent the statement that P ∈ AB ⇔ ¬P ∗ A ∗ B. P ∈ BA ⇔
−−→
¬P ∗ B ∗ A ⇔ ¬A ∗ B ∗ P (BAI) ⇒ ¬A ∗ C ∗ P (part(a)) ⇔ P ∈ CA .
−−−→ −−−→
(c) Suppose P ∈ BA ∩ BC . We must prove that P = B.
−−−→
−−→
−−→
Claim: P ∈ AC. Proof of claim: Since P ∈ BA , P ∈ CA by part (b). Similarly, P ∈ AC .
So by Proposition 3.1(a), P ∈ AC.
Claim: In fact, P ∈ AB ∩ BC = {B}, so obviously P = B. Proof: Since AC = AB ∪ BC
(Prop. 3.5), P ∈ AB or P ∈ BC. We claim that P ∈ AB. If not, then P ∈ BC, as we
just showed, but then an application of Proposition 3.3 (which I’ll leave to you), using the
−−−→
definition of BC, shows that P ∈
/ BA , contradicting the given. Similarly, P ∈ BC as well.
Thus P ∈ AB ∪ BC = {B}.
−−−→
−−−→
−−−→
5. What the first part of this problem shows is that, if P ∈ QR and P 6= Q, then QP = QR .
(Just use the definition of ray.) The trick to deducing that every ray has a unique opposite ray is
−−−→
to use this fact together with Proposition 3.4 (line separation). Given ray BA and opposite ray
−−−→
−−−→
←−→
−−−→
BC , consider any ray BP that emanates from B, is part of line BA , and is not equal to BA
−−−→
−−−→
−−−→
(definition of opposite ray). By Prop. 3.4, P ∈ BA or P ∈ BC . P ∈
/ BA , because if it were
−−−→ −−−→
−−−→
−−−→ −−−→
BP = BA . Thus P ∈ BC ; hence BP = BC .
6. (1) IAI (2) Prop. 2.3 (3) IAI (4) rest of BAII (5) BAI (6) Suppose G = D or D, B, and G
←−→
←−→
←−→
are collinear. In either case, BD = BG (IAI). Hence F lies on BD by step 5 (and IAI),
←−→ ←−→ ←−→
contradicting step (2). (7) rest of BAII (8) IAI (9) Suppose F = H. Then GB = GF = GH =
←−→
←−→
←−→
←−→
GD (IAI), contradicting step (6). (10) IAI (11) Lines GH = GD and F H are distinct, by
step 6, so they intersect only at G (which is distinct from D, also by step 6). (12) H does not
←−→
←−→
←−→
lie on BG , since by Proposition 2.1, distinct lines BG and GH intersect only at G (which is
←−→
←−→
distinct from H by step 7 and BAI). Therefore, BG and F H are distinct lines and intersect
only at F . (13) same reasoning as last step (14) steps 6 and 4. (15) As determined previously,
these lines are distinct, so Proposition 2.1 applies. (16) step 7 and BAIII (17) Pasch’s Theorem
together with steps 11, 12, 13, and 16 (18) step 17
1
−−→
9. Contrary to the suggestion in the text, I will use a direct proof. Let P ∈ AB , P 6= A. Then either
P = B, A ∗ P ∗ B, or A ∗ B ∗ P . In either of the latter two cases, since A is the unique point of
←−→
intersection of line l and line AB (Prop. 2.1), line l is not incident with a point of segment P B
(by definition of segment). Hence, P and B are on the same side of l by definition.
←−→
10. Assume D lies on line BC and is in the interior of ∠BAC. By definition of interior, D must lie
←−→
←−→
on the same side of AB as C, so in particular D does not lie on AB and cannot be equal to
B. Similarly, D 6= C. Thus, by BAIII, one of the following must hold: D ∗ B ∗ C, B ∗ D ∗ C, or
←−→
B ∗ C ∗ D. In the first case, D and C are on opposite sides of AB by definition, contradicting
the definition of interior. The last case is similarly eliminated, and we conclude that B ∗ D ∗ C.
Conversely, suppose B ∗ D ∗ C. By Proposition 2.1, B is the only point of intersection of lines
←−→
←−→
←−→
AB and BC . Hence, no point of segment DC lies on line AB , and D and C are on the same
←−→
←−→
side of AB . Similarly, B and D are on the same side of line AC . Thus, by definition, D is in
the interior of ∠BAC.
11. Parts (a) and (b) are easy exercises using similar arguments to the ones just demonstrated. For
part (c) I am aware of at least two proofs, one suggested by the hint in the text and the other
discovered in class. Both proofs rely on Proposition 3.7. It is not uncommon to have several
substantively different proofs of the same result, nor is it uncommon for different proofs to rely
on a common essential element. I discuss both proofs below.
Given E ∗ A ∗ C, we are to show that point B lies in the interior of ∠DAE. Since D lies in the
←−→ ←−→
interior of ∠BAC (given), B and D lie on the same side of line AC = AE . It remains to show
←−→
that E and B lie on the same side of line AD . Suppose, by way of contradiction, that they were
←−→
on opposite sides. Then a point P on line AD would be in segment EB.
←−→
−−→
Let F be a point such that F ∗ A ∗ D. By Proposition 3.4, any point on AD is either in ray AD
−−→
or its opposite ray, AF .
−−→
Case 1. Suppose P ∈ AD . Then by Proposition 3.7, P would be in the interior of ∠EAB; hence
D would be in the interior of ∠EAB, by Prop. 3.8 (a). (Here we have used the fact that, if
←−→
←−→
←−→
←−→
P ∈ AD , then AP = AD .) Then D and E would be on the same side of line AB , by the
←−→
definition of interior. It follows from the given that E and C are on opposite sides of AB (by
←−→
definition); hence, by the corollary to BAIV, C and D would be on opposite sides of AB . Thus,
D would not be in the interior of ∠BAC, contradicting the given.
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Case 2. On the other hand, suppose P ∈ AF . Then using similar arguments, which I leave to
←−→
you, it would follow that B and D are on opposite sides of line AC , again contradicting the given.
Since we reach a contradiction in either case, we conclude that E and B are on the same side of
←−→
←−→
line AD . Combining this result with the fact that B and D are on the same side of AE , proved
earlier, we have that B is in the interior of ∠EAD by definition.
←−→
Alternatively, consider segment ED. D and C are on the same side of AB . (Why?) C and E
are on opposite sides of this line. (Why?) Hence D and E are on opposite sides. (Why?) Thus a
←−→
point P on line AB is in segment ED.
2
−−→
We claim P ∈ AB , from which the desired result follows by Propositions 3.7 and 3.8(a). To prove
←−−→
the claim, note that P and D are on the same side of EAC (why?), while D and B are on the
same side (why?).
13. (a) The following lemma is convenient for this part: If a point is in the interior of two angles of
a triangle, then it is in the interior of the triangle. I leave the proof to you. You just need to
prove the point is in the interior of the third angle; use the definition of interior (of an angle).
Let P be a point in the exterior of 4ABC. Suppose ray r emanates from P and intersects
←−→
side AB at point Q. By Pasch’s Theorem, line P Q also intersects side AC or BC at a
point R. What we need to prove is that R is on ray P Q and not on its opposite ray. (By
Proposition 3.4, it must be on one or the other.)
By way of contradiction, suppose R ∗ P ∗ Q. We claim that P is in the interior of 4ABC,
contradicting the given.
Case 1. R = C. P is in the interior of angles ∠ABC and ∠BAC by Proposition 3.7, so P is
in the interior of the triangle by the lemma above.
Case 2. R 6= C. Without loss of generality, assume R is between A and C. (Why is there
”no loss of generality” in assuming this?) By Proposition 3.7, P is in the interior of ∠BAC.
←−→
By definition of interior, P is on the same side of line BA as R. It remains only to prove
←−→
that P and Q are on the same side of line BC . (Why is this all that remains to be proven?)
←−→
Point R is on the same side of BC as A. (Why?) Similarly, A and Q are on the same side
←−→
of BC , so Q and R are on the same side by BAIV. Suppose QP contained a point S on
←−→
line BC . By proposition 3.5, S is in segment QR; hence Q and R are on opposite sides of
←−→
←−→
BC , a contradiction. So QP contains no point lying on BC , and Q and P are on the same
←−→
side of BC by definition. (Note that this last argument is exactly the argument needed to
prove that a half-plane is convex. See Exercise 19. If we first proved the general result that
a half-plane is convex, we could apply that result here rather than repeat the argument used
to prove it.)
(b) For this part, the trick suggested by the text is invaluable. Following the text’s choice of
labels (see figure 3.38), let D be a point in the interior of 4ABC. Let r be a ray emanating
from D. Let l be the line of which r is a part.
−−→
By the crossbar theorem, ray AD intersects side BC in a point E. (The hypothesis of the
−−→
−−→
−−→
Crossbar Theorem requires that ray AD be between rays AB and AC . Why is this the
case?) Note that, since D is in the interior of the ∠C (why?), A ∗ D ∗ E by Proposition 3.7.
←−→
Case 1. l = AE . In this case, since A and E are on opposite rays emanating from D, one
of these points must be on ray r.
←−→
Case 2. l 6= AE . By Pasch’s Theorem, applied to 4BAE, l intersects either side AB or
BE at a point F . Applying Pasch’s Theorem to 4CAE, we also have that l intersects AC
or CE at a point G. The result follows from the following claim:
Claim: Either F or G belongs to ray r. To prove this claim, note that F and G cannot be
on the same side of 4ABC by Proposition 2.1, because they are distinct points. (Why?) So
F and G are on different sides of one angle of the triangle. Since D is in the interior of the
triangle, D must be in the interior of this angle, so F ∗ D ∗ G by Proposition 3.7. Now apply
an argument similar to the one used in Case 1.
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