Download Math 361 ACTIVITY 5: SSS Congruence theorem Why The SSS

Math 361
ACTIVITY 5: SSS Congruence theorem
Why
The SSS congruence criterion is the remaining general congruence theorem (leaving aside the special righttriangle situations). The proof illustrates the use of several of the plane separation and betweenness ideas, as
well as a general principle of ”copying a figure to a better position”.
LEARNING OBJECTIVES
1. Work as a team, using the team roles
2. Gain a deeper understanding of the role of plane separation in the axioms for Euclidean geometry
3. Gain skill in reasoning from the axioms in a formal mathematical development
CRITERIA
1. Success in completing the exercises.
2. Success in working as a team
RESOURCES
1. Your text - section 3.3
2. The typed notes on sections 3.3
3. 50 minutes
PLAN
1. Select roles, if you have not already done so, and decide how you will carry out steps 2 and 3 (5 minutes)
2. Work through the exercises given here - be sure everyone understands all results (30 minutes)
3. Assess the team’s work and roles performances and prepare the Reflector’s and Recorder’s reports including team
grade (5 minutes).
4. Be prepared to discuss your results.
EXERCISE In Neutral Geometry (axioms I-1 to I-2, D-1 to D-4, A-1 to A-3, H-1, and C-1), prove the SSS congruence
theorem as stated here by carrying out steps given here. You will need to be concerned about sides of lines and interiors
of angles as well as congruence of segments and of angles.
Theorem 1. If A, B, C are noncollinear points and P, Q, R are noncollinear points for which AB = P Q, AC = P R and
BC = QR then 4ABC ∼
= 4P QR
We will construct a congruent copy of 4P QR attached to 4ABC and prove that this copy is congruent to 4ABC
(with the right correspondence of vertices). Then we will have shown that both 4ABC and 4P QR are congruent to the
new triangle.
We start with the points and equal distances as given.
NOTE: You will (not “may”) want diagrams for keeping track of all the pieces – but you will need to realize that several
cases occur and one diagram cannot cover all cases.
1. By using the angle construction and segment construction theorems, construct a triangle 4AW C, with W on the
←→
opposite side of AC from B and satisfying 4AW C ∼
= 4P QR. (You need to prove your construction does give the
desired congruence).
NOTE: “Construct a triangle . . . ” means give instructions for building such a triangle and show the congruence
[based on your construction of the triangle]. Consider the examples in the proofs of Theorems 1 and 5.
With this congruence and the given data, we know [using the definition of congruence of triangles] that AW ∼
=
PQ ∼
= AB and CW ∼
= RQ ∼
= CB (You do not have to write a proof of this).
1
←→
−→
2. Prove [this is short – but important] segment W B meets AC at a point G with W − G − B (This will make W G =
−→
−→
−→
W B and BG = BW — thats our ray renaming theorem — which will be important for naming some angles).
←→
Now we consider: Where is G on AC? There are 5 possibilities - either G is one of the two points we’ve used to
name the line (G = A or G = C) or one of A, G, B is between the other two. Which case occurs will depend on the
shape of the triangles — in particular, whether various angles are acute, obtuse, or right angles.
3. Draw three figures, illustrating the cases a.) G = A, b.) A − G − C c.) A − C − G [you do not need to illustrate
the other two cases – case G = C looks like case a, case G − A − C looks like case c but with names switched]
Check (identify the necessary triangles and pieces) that the conditions of Lemma A hold for 4ABW (in the cases
where there is such a triangle) and for 4CBW (in the cases where there is such a triangle), so that we always have
6 AW B ∼
= 6 ABW and 6 CW B ∼
= 6 CBW if these angles all exist (Note that one check will cover all cases)
4. Prove:(This is case a.)) If G = A (that is, if A, B, W are collinear, so there is no triangle 4ABW , no angle 6 AW B
and no angle 6 ABW ) then 6 AW C ∼
= 6 ABC. (Consider angle renaming)
How would this proof be changed to cover the case G = C [result would still be 6 AW C ∼
= 6 ABC]? (These two
situations cover all the cases in #3 for which some of the names don’t really give triangles and angles)
5. Prove: (This is case b.)) If A − G − C then 6 AW C ∼
= 6 ABC.
(This is a bit trickier — you need to use the fact that G is between A and C to allow combining measures of
appropriate angles in the appropriate way)
6. Prove: (This is case c.)) If A − C − G then 6 AW C ∼
= 6 ABC (You need to use the fact that C is between G and A
to allow combining measures of appropriate angles in the appropriate way. This works a lot like the previous step,
but in an “opposite” direction ) How would this proof be changed to cover the case G − A − C?
7. (You don’t need to prove anything further). We have now shown that, in every case, AB ∼
= AW , 6 ABC ∼
=
∼
∼
6 AW C and W C ∼
= 4ABC .
= BC so Axiom C-1 tells us that 4AW C = 4ABC and we have 4P QR = 4AW C ∼
This completes he proof of the SSS criterion for congruence of triangles
SKILL EXERCISES: (hand in - individually - with this week’s assignments)
Read Section 3.4 (exterior angle theorem)
Write 1. p. 148 (section 3.3) # 18(Use the “between” relations that are shown in the diagram), 20 [Prove as
“If the perpendicular bisectors of two sides of a triangle meet, their intersection point is equidistant from all
three vertices and the perpendicular bisector of the third side also contains this point” You need to provide
the names and state the result clearly.]
2