Download Absolute geometry

Absolute geometry
Congruent triangles - SAS, ASA, SSS
Material for this section references College Geometry: A Discovery Approach, 2/e, David C. Kay, Addison Wesley, 2001. In particular, see section 3.3, pp 139-150. The
problems are all from section 3.3.
Proving ASA from SAS
ASA Theorem
If, under some correspondence, two angles and the included side of one triangle are congruent to the
corresponding angles and included side of another, then the triangles are congruent under that correspondence.
[Kay, p 141]
picture
conclusions
justifications
Suppose you have two triangles which
satisfy the hypothesis of ASA: two
angles and included side are congruent to the corresponding angles and
side.
Hypothesis (Given)
The goal is to show that having ASA (thing you want to prove) always leads to having SAS (the postulate); i.e.
having the corresponding parts congruent that are marked below:
always leads to having the additional marked sides congruent:
which proves the triangles are congruent by SAS.
So, prove that AC ∼
= XZ, given ∠ A ∼
= ∠ X, AB ∼
= XY , ∠ B ∼
= ∠ Y . This is done using a proof by contradiction.
picture
conclusions
∠A ∼
= ∠ X,
justifications
Given
AB ∼
= XY ,
∠B ∼
= ∠Y.
AC ∼
XZ.
=
Since they are not congruent,
there are two possibilities: either
AC > XZ or AC < XZ.
Suppose AC > XZ.
We can place a point D on AC with
A − D − C, and AD = XZ.
Therefore, ABD ∼
= XY Z.
Since the triangles are congruent, we
must have ∠ ABD ∼
= ∠ XY Z
Assumption. In a proof by contradiction, one should always assume:
picture
conclusions
justifications
Since A − D − C, D is in the interior
−→ −→ −→
of ∠ B, and BA − BD − BC
This is from Theorem 3 on page 108
(with definition of beteenness for rays
thrown in). If I were going to cite
this in a non-textbook specific way,
I’d call it an application of the Crossbar Theorem.
−→ −→ −→
Since BA − BD − BC,
m∠ ABD + m∠ DBC = m∠ ABC.
However, we were given that
∠ ABC ∼
= ∠ XY Z, and have shown
that ∠ ABD ∼
= ∠ XY Z. Therefore
m∠ XY Z + m∠ DBC = m∠ XY Z,
so m∠ XY Z < m∠ XY Z.
Contradiction.
It’s not over yet - that was one case of the “or”. The proof of the other case is similar; provide both conclusions and
justifications:
picture
conclusions
justifications
Suppose AC < XZ.
Case 2 of AC > XZ or AC < XZ
We can extend AC so A − C − D, and
AD = XZ.
picture
conclusions
justifications
Since A − C − D,[...]
However, we were given that
∠ ABC ∼
= ∠ XY Z,
and have shown that
Contradiction.
Therefore, it is not possible that AC ∼
XZ; we must have AC ∼
=
= XZ, and ABC ∼
= XY Z by the SAS Postulate.
Which finally gives us the ASA Theorem; assuming ASA always forces SAS, so ASA is sufficient to show triangles
congruent.