McGraw Hill Glencoe Geometry, 2012
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McGraw Hill Glencoe Geometry, 2012 View details
4. Parallel Lines and Proportional Parts
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Exercise 32 Page 497

Look for a common angle to both triangles. Use the Segment Addition Postulate to show that the corresponding sides that include the common angle are proportional. Also, use the Side-Angle-Side (SAS) Similarity Theorem to obtain the desired results.

Statements
Reasons
1.
△ FGH, J and K are midpoints of FH and HG respectively
1.
Given
2.
FH = FJ+JH and GH = GK+KH
2.
Segment Addition Postulate
3.
FJ = JH and GK = KH
3.
Definition of congruent segments
4.
FH = 2HJ and GH = 2HK
4.
Substitution
5.
FH/HJ = 2 and GH/HK = 2
5.
Division Property of Equality
6.
FH/HJ = GH/HK
6.
Transitive Property of Equality
7.
∠ H ≅ ∠ H
7.
Reflexive Property of Congruence
8.
△ FGH ~ △ JKH
8.
SAS Similarity Theorem
9.
∠ F ≅ ∠ HJK
9.
Definition of similar triangles
10.
JK∥FG
10.
Corresponding Angles Converse
11.
FG/JK = GH/KH
11.
Definition of similar triangles
12.
FG/JK = 2
12.
Substitution
13.
JK=1/2FG
13.
Simplifying
Practice makes perfect

Let's consider △ FGH and let J and K be the midpoints of FH and GH, respectively.

Our mission is to prove that JK∥ FG and that JK = 12FG. We will begin to map out these relationships by using the Segment Addition Postulate to write the following pair of equations. FH = FJ+JH GH = GK+KH We can break down this equation further by checking for the relationships expressed on the diagram. By the definition of a midpoint, we know that FJ≅JH and GK≅KH. These relationships imply that FJ=JH and GK=KH, respectively.

FH = 2HJ GH = 2HK ⇒ FH/HJ = 2 [0.25cm] GH/HK = 2 Because both proportions equal the same number 2, we can use the Transitive Property of Equality to obtain the following equation. FH/HJ = GH/HK Additionally, notice that ∠ H is common to both △ FGH and △ JKH. The Reflexive Property of Congruence gives us ∠ H ≅ ∠ H. FH/HJ = GH/HK and ∠ H ≅ ∠ H In consequence, the Side-Angle-Side (SAS) Similarity Theorem tells us that △ FGH ~ △ JKH. Therefore, the corresponding angles are congruent.

The Corresponding Angles Converse gives us that JK∥FG. Additionally, the similarity relation gives us the proportions below. FG/JK = GH/KH = FH/JH Finally, let's solve the left-hand side equation for JK.
FG/JK = GH/KH
FG/JK = 2
Solve for JK
FG = 2JK
1/2FG = JK
JK = 1/2FG
We have proven the Triangle Midsegment Theorem.

Two-Column Proof

Given: & △ FGH, J andK are midpoints of & FHandHG respectively Prove: & JK∥FGandJK = 12FG Let's summarize the proof we did above in the following two-column table.

Statements
Reasons
1.
△ FGH, J and K are midpoints of FH and HG respectively
1.
Given
2.
FH = FJ+JH and GH = GK+KH
2.
Segment Addition Postulate
3.
FJ = JH and GK = KH
3.
Definition of congruent segments
4.
FH = 2HJ and GH = 2HK
4.
Substitution
5.
FH/HJ = 2 and GH/HK = 2
5.
Division Property of Equality
6.
FH/HJ = GH/HK
6.
Transitive Property of Equality
7.
∠ H ≅ ∠ H
7.
Reflexive Property of Congruence
8.
△ FGH ~ △ JKH
8.
SAS Similarity Theorem
9.
∠ F ≅ ∠ HJK
9.
Definition of similar triangles
10.
JK∥FG
10.
Corresponding Angles Converse
11.
FG/JK = GH/KH
11.
Definition of similar triangles
12.
FG/JK = 2
12.
Substitution
13.
JK=1/2FG
13.
Simplifying