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 (fill in and submit as GR5 – part 1)
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
Suppose you have two triangles
which satisfy the hypothesis of
ASA: two angles and included
side are congruent to the
corresponding angles and side.
JUSTIFICATIONS
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 then allows us to conclude the triangles are congruent by the SAS Postulate (which we assume as
the basis for this geometry). This establishes ASA as a theorem; if we start with the ASA hypothesis
(“two angles and the included side of one triangle are congruent to the corresponding angles and
included side of another”), then the conclusion is the triangles must be congruent.
So, we need to prove that AC  XZ , given A  X , AB  XY , B  Y . This is done using a
proof by contradiction.
PICTURE
CONCLUSIONS
A   X
AB  XY
B   Y
JUSTIFICATIONS
Hypothesis (Given)
Since they are not
congruent, there are two
possibilities: either
AC  XZ or AC  XZ
.
Assumption. In a proof
by contradiction, one
should always assume:
[Fill in 1]
Suppose AC  XZ .
[Fill in 2]
We can place a point D
[Fill in 3]
on AC with A  D  C ,
and AD  XZ .
Therefore,
ABD  XYZ .
[Fill in 4]
Since the triangles are
congruent, we must have
ABD  XYZ .
[Fill in 5]
Since A  D  C , D is
in the interior of  B ,
This is from Theorem 3
on page 108 (with
definition of
betweenness 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.
and BA  BD  BC .
Since BA  BD  BC ,
[Fill in 6]
mABD  mDBC
 mABC
.
However, we were given
that ABC  XYZ ,
and have shown that
ABD  XYZ .
Therefore
[Fill in 7]
mXYZ  mDBC
 mXYZ
and so
mXYZ  mXYZ .
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
Now suppose AC  XZ
.
JUSTIFICATIONS
Case 2 of AC  XZ or
AC  XZ .
We can extend AC so
A  C  D , and
AD  XZ .
[Fill in 8]
[Fill in 9]
[Fill in 10]
[Fill in 11]
[Fill in 12]
Since A  C  D ,
[Fill in 13]
[Fill in 14]
[Fill in 15]
[Fill in 16]
[Fill in 17]
[Fill in 18]
Therefore, it is not possible that AC  XZ ; we must have AC  XZ , and
Postulate.
ABC  XYZ by the SAS
Which finally gives us the ASA Theorem; if we assume ASA, we get to SAS, so ASA is sufficient to show
triangles congruent.