McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
5. Isosceles and Equilateral Triangles
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Exercise 42 Page 381

Consider that vertical angles are congruent.

m∠ JLK=20

Practice makes perfect

In a previous exercise, we found that m∠ LPM=m∠ LMP=80.

We want to find the measure of the angle JLK. To do so, we will first find the measure of the angle MLP by using the Triangle Sum Theorem.
m∠ MLP+m∠ LMP+m∠ LPM=180
m∠ MLP+80+80=180
m∠ MLP+160=180
m∠ MLP=20
Now that we found m∠ MLP, we can find m∠ JLK. Notice that ∠ JLK and ∠ MLP are vertical angles, which means that they are congruent. Then, m∠ JLK=m∠ MLP.
m∠ JKL=m∠ MLP
m∠ JKL=20
Therefore, m∠ JLK is 20.

Extra

Vertical Angles Theorem

In this solution, we used the fact that vertical angles are always congruent. This is true due the Vertical Angle Theorem. We will learn a bit more about this theorem by considering the next diagram.

Two intersecting lines that form two pairs of vertical angles

Analyzing the diagram, it can be observed that ∠ 1 and ∠ 2 form a straight angle, so these are supplementary angles. Similarly, ∠ 2 and ∠ 3 are also supplementary angles.

Two intersecting lines that form two pairs of vertical angles

Therefore, by the Angle Addition Postulate, the sum of m∠ 1 and m∠ 2 is 180^(∘), and the sum of m∠ 2 and m∠ 3 is also 180^(∘). These facts can be used to express m∠ 2 in terms of m∠ 1 and also in terms of m∠ 3.

Angle Addition Postulate Isolate m∠ 2
m∠ 1+m∠ 2 = 180^(∘) m∠ 2 = 180^(∘)-m∠ 1
m∠ 2+m∠ 3 = 180^(∘) m∠ 2 = 180^(∘)-m∠ 3
By the Transitive Property of Equality, the expressions representing m∠ 2 can be set equal to each other. m∠ 2= 180^(∘)-m∠ 1 m∠ 2= 180^(∘)-m∠ 3 ⇓ 180^(∘)-m∠ 1= 180^(∘)-m∠ 3 Then the obtained equation can be simplified.
180^(∘)-m∠ 1=180^(∘)-m∠ 3
- m∠ 1=- m∠ 3
m∠ 1=m∠ 3
By the definition of congruent angles, this means that the vertical angles ∠ 1 and ∠ 3 are congruent angles. Using the same argumentation, ∠ 2 and ∠ 4 can also be proven to be congruent.