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Chapter 4.2 Notes: Apply
Congruence and Triangles
Goal: You will identify congruent figures.
• Two geometric figures are congruent if they have
exactly the same size and shape.
Congruence Statements
• In two congruent figures, all the parts of one figure
are congruent to the corresponding parts of the
other figure.
• In congruent polygons, this means that the
corresponding sides and the corresponding angles
are congruent.
Identifying congruent figures
• Two geometric figures are congruent if they have exactly the
same size and shape.
NOT CONGRUENT
CONGRUENT
Triangles
Corresponding angles
A ≅ P
B ≅ Q
C ≅ R
Corresponding Sides
AB ≅ PQ
BC ≅ QR
CA ≅ RP
B
A
Q
CP
R
Ex.1: Write a congruence statement for the triangles.
Identify all pairs of congruent corresponding parts.
Ex.2: In the diagram, DEFG  SPQR .
a. Find the value of x.
b. Find the value of y.
Ex.3: In the diagram below, ABGH  CDEF .
a. Identify all pairs of congruent corresponding
parts.
b. Find the value of x and find mH.
Ex.4: Show that PTS  RTQ.
• Theorem 4.3 Third Angles Theorem:
If two angles of one triangle are congruent to two
angles of another triangle, then the third angles are
also congruent.
Ex. 3 Using the Third Angles Theorem
• Find the value of x.
M
R
55°
N
65°
L
• In the diagram, N ≅ R
and L ≅ S. From the
Third Angles Theorem, you
(2x + 30)° T know that M ≅ T. So
mM = mT. From the
Triangle Sum Theorem,
mM=180° - 55° - 65° =
60°
• mM = mT
60° = (2x + 30)°
S
30 = 2x
15 = x
Ex.5: Find mBDC .
Ex.6: In the diagram, what is mDCN ?
Ex.7: The two triangles are congruent. Identify all
angles and sides that are congruent. Then write a
congruence statement.
Ex.8: In the diagram, ABCD  FGHK.
a. Find the value of x.
b. Find the value of y.
Properties of Congruent Triangles
• Theorem 4.4 Properties of Congruent Triangles:
– Reflexive Property of Congruent Triangles:
For any triangle ABC, ABC  ABC .
– Symmetric Property of Congruent Triangles:
If ABC  DEF , then DEF  ABC .
– Transitive Property of Congruent Triangles:
If ABC  DEF & DEF  JKL , then
ABC  JKL .
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