Exercise 8 Page 587

Practice makes perfect

The pentagon ABCDE is inscribed in a circle. Let's make a diagram describing the situation.

A figure consisting of a circle with a red pentagon inscribed within it. The vertices of the pentagon are marked with the letters A, B, C, D, and E in counterclockwise order, starting from the top vertex.

Regular polygons have congruent sides and congruent angles. Let's use the Polygon Interior Angles Theorem to find the sum of the angle measures of the pentagon.

(n-2)* 180^(∘) ⇒ (5-2)* 180^(∘)=540^(∘) By dividing the sum of the angle measures by 5 we find the measure of each angle. 540^(∘)/5=108^(∘) Thus, each interior angle in the pentagon has a measure of 108^(∘).

A regular pentagon with angle measures inscribed in a circle

We want to describe the relationship between ∠ CDE and ∠ CAE. Let's view these angles in the diagram.

Since ∠ CDE is one of the interior angles of the pentagon we have already calculated its measure. m∠ CDE = 108^(∘) The segment AC divides the pentagon into a quadrilateral and an isosceles triangle. By the Base Angles Theorem the triangle's base angles are congruent. Knowing this and by using the Triangle Sum Theorem, we can find their measure.

m∠ ABC +m∠ BCA + m∠ CAB =180

Substitute

m∠ ABC= 108

108 +m∠ BCA + m∠ CAB =180

Substitute

m∠ BCA= m∠ CAB

108 + m∠ CAB + m∠ CAB =180

▼

Solve for m∠ CAB

AddTerms

Add terms

108 +2(m∠ CAB) =180

SubEqn

LHS-108=RHS-108

2(m∠ CAB) =72

DivEqn

.LHS /2.=.RHS /2.

m∠ CAB =36

Let's use that m∠ CAB and m∠ CAE add to 108. 36 + m∠ CAE=108 ⇔ m∠ CAE=72 We will now factor the angle measures to help us establish a relationship between them. m∠ CAE=72 m∠ CDE=108 ⇕ m∠ CAE=2* 36 m∠ CDE=3 * 36 ⇕ m∠ CAE * 3/2=m∠ CDE

Again we need to consider a diagram that describes pentagon ABCDE, which is inscribed in a circle.

This time we want to find a relationship between ∠ CBE and ∠ CAE.

When we create ∠ CBE we draw the segment BE. This segment divides the pentagon into a quadrilateral and an isosceles triangle. By the Base Angles Theorem the triangle's base angles are congruent. Knowing this and by using the Triangle Sum Theorem, we can find their measure.

m∠ BAE +m∠ AEB + m∠ EBA =180

Substitute

m∠ BAE= 108

108 +m∠ AEB + m∠ EBA =180

Substitute

m∠ AEB= m∠ EBA

108 + m∠ EBA + m∠ EBA =180

▼

Solve for m∠ EBA

AddTerms

Add terms

108 +2(m∠ EBA) =180

SubEqn

LHS-108=RHS-108

2(m∠ EBA) =72

DivEqn

.LHS /2.=.RHS /2.

m∠ EBA =36

Let's now use that m∠ EBA and m∠ CBE add to 108. 36 + m∠ CAE=108 ⇔ m∠ CBE=72 Let's now study ∠ CAE. When we create this angle we need to draw a segment which divides the pentagon into a quadrilateral and an isosceles triangle. The triangle, △ CBA, has two sides and the included angle congruent with △ BAE. By the Side-Angle-Side Congruence Theorem the triangles are congruent. △ CBA ≅ △ BAE By the definition of congruence we know that both triangles' base angles have the same measure. m∠ BAC = m∠ EBA = 36 Let's use that m∠ BAC and m∠ CAE add to 108. 36 + m∠ CAE=108 ⇔ m∠ CAE=72 We have found that ∠ CBE and ∠ CAE have the same measure.

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Exercise 8 Page 587

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