Exercise 2 Page 253

Practice makes perfect

a Consider the given diagram.

Let's now recall the Isosceles Triangle Theorem.

Isosceles Triangle Theorem
If two sides of a triangle are congruent, then the angles opposite those sides are congruent.

Since we know that DE ≅ EF, we know by the theorem that ∠ D ≅ ∠ F. Therefore, these two angles have the same measure.

Next, we know by the Triangle Angle-Sum Theorem that the sum of the measures of the interior angles of a triangle equals 180^(∘).

m ∠ A+m ∠ B+m ∠ C=180 Applying the theorem, we can write an equation in terms of x. x+x+30=180 Let's solve the equation for x.

x+x+30=180

▼

Solve for x

AddTerms

Add terms

2x+30=180

SubEqn

LHS-30=RHS-30

2x=150

DivEqn

.LHS /2.=.RHS /2.

x=75

b Consider the given diagram. Note that NL ≅ NM and therefore △ LNM is an isosceles triangle whose vertex angle is ∠ LNM. Moreover, this vertex angle is bisected by NO, which means that m∠ LNO = x^(∘).

Let's recall a theorem that can be applied in this situation.

If a line bisects the vertex angle of an isosceles triangle, then the line is also the perpendicular bisector of the base.

We can say, by this theorem, that ON is the perpendicular bisector of ∠ LNM. Therefore, LM and ON are perpendicular and ∠LON is a right angle.

Next, we apply use the Triangle Angle-Sum Theorem on △ LON.

x+42+90=180

AddTerms

Add terms

x+132=180

SubEqn

LHS-132=RHS-132

x=48

Subchapter links

Lesson Check p.253

Standardized Test Prep p.256

Mixed Review p.256

Exercise 2 Page 253

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