Exercise 13 Page 467

Let's focus on the part of the figure that might be relevant to answer the question.

• The question asks for the measure of an angle, so we concentrate on angle related information and ignore the length measurements.
• We are asked to find m∠ DHG, so let's concentrate on quadrilateral DHGB. The segment HB cuts this into two triangles.
• The Incenter Theorem tells us that the incenter of a triangle is the point of concurrency of the angle bisectors. Let's add angle markers at B indicating the congruent angles ∠ DBH and ∠ GBH.

We will first use the blue right triangle to find m∠ DHB, and the green right triangle to find m∠ GHB.

Finding m∠ DHB.

According to Corollary 4.1 of the Triangle Angle-Sum Theorem, the acute angles of a right triangle are complementary. Since it is given that m∠ DBH=30, this will let us find m∠ DHB.

m∠ DBH+m∠ DHB=90

Substitute

m∠ DBH= 30

30+m∠ DHB=90

SubEqn

LHS-30=RHS-30

m∠ DHB=60

Finding m∠ GHB.

The blue and green triangles have two congruent angles. According to the Third Angles Theorem the third angles are also congruent, so they have the same measure. m∠ GHB=m∠ DHB = 60

Answering the question

We now know the measure of both parts of angle ∠ DHG. We can use the Angle Addition Postulate to find m∠ DHG as the sum of these two measures.

m∠ DHG=m∠ DHB+m∠ GHB

SubstituteII

m∠ DHB= 60, m∠ GHB= 60

m∠ DHG= 60+ 60

AddTerms

Add terms

m∠ DHG=120

The measure of ∠ DHG is m∠ DHG=120.