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Section 4.3Congruent Triangles
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If two geometric figures have exactly the same shape and size, they are congruent.
In two congruent polygons, all of the parts of one polygon are congruent to the corresponding parts or matching parts of the other polygon. These corresponding parts include corresponding angles and corresponding sides.
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Concept 1
Other congruence statements for the triangles above exist. Valid congruence statements for congruent polygons list corresponding vertices in the same order.
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Example 1:
a) Show that the polygons are congruent by identifying all of the congruent corresponding parts. Then write a congruence statement.
Angles: A R, B T, C P, D S, E Q
Sides: , , , ,AB RT BC TP CD PS DE SQ EA QR
All corresponding parts of the two polygons are congruent. Therefore, ABCDE RTPSQ.
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b) The support beams on the fence form congruent triangles. In the figure ΔABC ΔDEF, which of the following congruence statements correctly identifies corresponding angles or sides?
a)
b)
c)
d)
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The phrase “if and only if” in the congruent polygon definition means that both the conditional and converse are true. So, if two polygons are congruent, then their corresponding parts are congruent.
For triangles we say Corresponding Parts of Congruent Triangles are Congruent or CPCTC.
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Example 2:
a) In the diagram, ΔITP ΔNGO. Find the values of x and y.
O PCPCTCmO = mP Definition of congruence6y – 14= 40 Substitution
6y = 54 Add 14 to each sidey = 9 Divide each side by 6
NG = IT Definition of congruencex – 2y = 7.5 Substitutionx – 2(9) = 7.5 y = 9x – 18 = 7.5 Simplifyx = 25.5 Add 18 to each side
CPCTCNG IT
x = 25.5, y = 9
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b) In the diagram, ΔFHJ ΔHFG. Find the values of x and y.
GFH HFG so:6x + 8 = 35, solve for x.x = 4.5
so:
2 3 2.5, solve for .
2.75
FJ HG
FJ HG
y y
y
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Concept 2
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Example 3: ARCHITECTURE A drawing of a tower’s roof is composed of congruent triangles all converging at a point at the top. If IJK IKJ and mIJK = 72, find mJIH.
mIJK + mIKJ + mJIK = 180° Triangle Angle-Sum TheoremΔJIK ΔJIH Congruent Triangles
mIJK + mIJK + mJIK = 180° Substitution72° + 72° + mJIK = 180° Substitution
144° + mJIK = 180° SimplifymJIK = 36° Subtract 144
from each sidemJIH = 36° Third Angles
Theorem
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Example 4: Write a two-column proof.
Prove: ΔLMN ΔPON
Given: ÐL @ ÐPLM POLN PNMN NO
Statements Reasons
1. 1.
2. 2.
3. 3.
4. 4.
5. 5.
6. 6.
7. 7.
ÐL @ ÐP Given
LM PO
LN PN
MN NO
Given
Given
Given
LNM PNO
M OΔLMN ΔPON
Vertical Angles Theorem
Third Angles Theorem
CPCTC
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Concept 3
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