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TEKS Focus:
(6)(B) Prove two triangles are congruent by applying
the Side-Angle-Side, Angle-Side-Angle, Side-SideSide, Angle-Angle-Side, and Hypotenuse-Leg
congruence conditions.
(1)(F) Analyze mathematical relationships to
connect and communicate mathematical ideas.
(1)(A) Apply mathematics to problems arising in
everyday life, society, and the workplace.
(1)(D) Communicate mathematical ideas, reasoning,
and their implications using multiple
representations, including symbols, diagrams,
graphs, and language as appropriate.
(1)(G) Display, explain, or justify mathematical ideas
and arguments using precise mathematical
language in written or oral communication.
An included side is the common side of
two consecutive angles in a polygon.
The following postulate uses the idea of
an included side.
4-3
Determine if you can use ASA to prove the triangles
congruent. Explain.
Two congruent angle pairs are given, but the congruent
side (by the Reflexive Property of congruence) is NOT
the included side for the given congruent angles.
Therefore ASA cannot be used to prove the triangles
congruent.
You can use the Third Angles Theorem to prove another congruence
relationship based on ASA. This theorem is Angle-Angle-Side (AAS).
4-2
Determine if you can use ASA to
prove NKL  LMN. Explain.
By the Alternate Interior Angles Theorem. KLN  MNL.
NL  LN by the Reflexive Property. No other congruence
relationships can be determined, so ASA cannot be
applied.
Determine if you can use AAS to prove the triangles
congruent with the given information. Explain.
Given: LJ bisects KLM, K  M
K  M, and KLJ  MLJ by the definition of bisectors.
LJ  LJ by the reflexive property of congruence. Therefore
we can use AAS to prove the triangles congruent.
ΔHOG  ΔATC by ASA.
(Note: FN is not the included side in ΔNFI.
There is a given pair of congruent angles;
a pair of vertical angles congruent,
and PR and SR are the non-included sides.
ΔPAR  ΔSIR by AAS.
3x – 10 = 2x + 20
x – 10 = 20
x = 30
ABC  DCB by Alt. Int. s Thm.
BC  BC by Reflexive Property of 
If x = 30, then ACB  CBD.
ΔABC  ΔDCB by ASA.
B
A 3
Given:
3
4
AC  EC
Prove: ABC  EDC
C
3  4
2.
AC  EC
3.
1  2
4.
ABC  EDC
2
4
D
STATEMENTS
1.
1
E
REASONS
1.
2.
3.
4.
Given
Given
Vertical s Thm.
ASA
Given: AB || DE
BC  DC
Prove: ABC  EDC
B
A
C
E
D
STATEMENTS
1.
REASONS
1.
AB // DE
Given
2.
2.
A  E
Alt. Int. s Thm.
3. ACB  ECD 3.Vertical s Thm.
4.
4.
BC  DC
Given
5. ABC  EDC 5. AAS