Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
2. Section 7.2
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Exercise 73 Page 401

Use the given information to prove that two angles and the included side of the triangles are congruent.

See solution.

Practice makes perfect

Let's add the given information that BC∥ EF, AB∥ DE, and AF=DC.

If we view FC as a transversal to BC and EF, then ∠ BCF and ∠ EFC are alternate interior angles. We can make the same argument for ∠ BAC and ∠ EDF if we view AD as a transversal to AB and DE. Since the two lines cut by the third line are parallel, we can by the Alternate Interior Angles Theorem claim that they are congruent.

In △ ABC and △ DEF we see that their horizontal sides share FC. Therefore, FC=FC according to the Reflexive Property of Congruence.

Now we have enough information to claim that AC=DF. If we separate the triangles, we see that two pairs of angles and the included side are congruent.

With this information we can claim congruence by the ASA Congruence Theorem. As seen in the diagram, BC and EF are corresponding sides, which means they are congruent. Now we can complete the table.

Statements
Reasons
1.
BC∥ EF
1.
Given
2.
m∠ BCF = m∠ EFC
2.
Alternate Interior Angles Theorem
3.
AB∥ DE
3.
Given
4.
m∠ BAC = m∠ EDF
4.
Alternate Interior Angles Theorem
5.
AF=DC
5.
Given
6.
FC=FC
6.
Reflexive Property of Equality
7.
AF+FC=FC+DC
7.
Additive Property of Equality
8.
AC=DF
8.
Segment addition
9.
△ ABC ≅ △ DEF
9.
ASA Congruence Theorem
10.
BC ≅ EF
10.
≅ Δ s → ≅ parts