## Engage NY Eureka Math Geometry Module 5 Lesson 6 Answer Key

### Eureka Math Geometry Module 5 Lesson 6 Exercise Answer Key

Opening Exercise

In a circle, a chord \(\overline{D E}\) and a diameter \(\overline{A B}\) are extended outside of the circle to meet at point C. If m∠DAE = 46°, and m∠DCA = 32°, find m∠DEA.

Let m∠DEA = y˚, m∠EAB = x˚

In △ABD, m∠DBA =

Reason:

Answer:

In △ABD, m∠DBA = y˚

Reason: angles inscribed in same arc are congruent

m∠ADB =

Reason:

Answer:

m∠ADB = 90° Reason: angle inscribed in semicircle

∴46 + x + y + 90 =

Reason:

Answer:

∴46 + x + y + 90 = 180 Reason: sum of angles of triangle is 180°

x + y =

Answer:

x + y = 44

In △ACE, y = x + 32

Reason:

Answer:

In △ACE, y = x + 32

Reason: Exterior angle of a triangle is equal to the sum of the remote interior angles

x + x + 32 =

Reason:

Answer:

x + x + 32 = 44

Reason: substitution

x =

Answer:

x = 6

y =

Answer:

y = 38

m∠DEA =

Answer:

m∠DEA = 38°

Exercises

Find the value of x in each figure below, and describe how you arrived at the answer.

Exercise 1.

Hint: Thales’ theorem

Answer:

m∠BEC = 90° inscribed in a semicircle

m∠EBC = m∠ECB = 45° base angles of an isosceles triangle are congruent and sum of angles of a triangle = 180°

m∠EBC = m∠EDC = 45° angles inscribed in the same arc are congruent

x = 45

Exercise 2.

Answer:

m∠BAD = 146˚, if parallel lines cut by a transversal, then interior angles on the same side are supplementary. Then the \(m \widehat{B D}\) = 146˚, because ∠BAD is a central angle intercepting \(\widehat{B D}\). Then remaining arc of the circle, \(\widehat{B C D}\), has a measure of 214°. Then m∠BED = 107˚ since it is an inscribed angle intercepting \(\widehat{B C D}\). The angle sum of a quadrilateral is 360°, which means x = 73.

Exercise 3.

Answer:

m∠BEC = m∠CFB = \(\frac{1}{2}\) m∠BAC = 52°

Inscribed angles are half the measure of the central angle intercepting the same arc.

m∠DEG = 128° linear pair with ∠BEC

m∠GFD = 128° linear pair with ∠CFB

m∠EGF = 74° sum of angles of a quadrilateral

x = 74 vertical angles

Exercise 4.

Answer:

The measures of arcs \(\widehat{D E}\), \(\widehat{E F}\), and \(\widehat{F C}\) are each 60˚, since the intercepted arc of an inscribed angle is double the measure of the angle. This means \(\widehat{m D E} C\) = 180˚, or \(\widehat{D E C}\) is a semicircle. This means x is 90, since ∠DBC is inscribed in a semicircle.

### Eureka Math Geometry Module 5 Lesson 6 Problem Set Answer Key

In Problems 1–5, find the value x.

Question 1.

Answer:

x = 40.5

Question 2.

Answer:

x = 57

Question 3.

Answer:

x = 15

Question 4.

Answer:

x = 34

Question 5.

Answer:

x = 90

Question 6.

If BF = FC, express y in terms of x.

Answer:

y = 90 – \(\frac{x}{2}\)

Question 7.

a. Find the value of x.

Answer:

x = 90

b. Suppose the m∠C = a°. Prove that m∠DEB = 3a°.

Answer:

m∠D = a° (alternate angles are equal in measure), m∠A = 2a° (inscribed angles half the central angle), a° + 2a° + m∠AED = 180° (the sum of the angles of triangle is 180°), m∠AED = (180 – 3a)°, m∠AED + m∠DEB = 180° (angles form line), (180 – 3a)° + m∠DEB = 180° (substitution), m∠DEB = 3a°

Question 8.

In the figure below, three identical circles meet at B, F, C, and E, respectively. BF = CE. A, B, C and F, E, D lie on straight lines.

Prove ACDF is a parallelogram.

PROOF:

Join BE and CF.

BF = CE Reason: ______________________________

a = __________ = __________ = __________ = d Reason: ______________________________

__________ = __________ Alternate interior angles are equal in measure.

\(\overline{A C}\) ∥ \(\overline{F D}\)

__________ = __________ Corresponding angles are equal in measure.

\(\overline{A F}\) ∥ \(\overline{B E}\)

__________ = __________ Corresponding angles are equal in measure.

\(\overline{B E}\) ∥ \(\overline{C D}\)

\(\overline{A F}\) ∥ \(\overline{B E}\)∥\(\overline{C D}\)

ACDF is a parallelogram.

Answer:

Join BE and CF.

BF = CE Reason: Given

a = b = f = e = d Reason: Angles inscribed in congruent arcs are equal in

m∠CBE = m∠FEB Alternate interior angles are equal in measure.

\(\overline{A C}\) ∥ \(\overline{F D}\)

m∠A = m∠CBE Corresponding angles are equal in measure.

\(\overline{A F}\) ∥ \(\overline{B E}\)

m∠D = m∠BEF Corresponding angles are equal in measure.

\(\overline{B E}\) ∥ \(\overline{C D}\)

\(\overline{A F}\) ∥ \(\overline{B E}\)∥\(\overline{C D}\)

ACDF is a parallelogram.

### Eureka Math Geometry Module 5 Lesson 6 Exit Ticket Answer Key

Question 1.

Find the measure of angles x and y. Explain the relationships and theorems used.

Answer:

m∠EAC = 42° (linear pair with ∠BAE). m∠EFC = \(\frac{1}{2}\) m∠EAC = 21° (inscribed angle is half measure of central angle with same intercepted arc). x = 21.

m∠ABD = m∠EAC = 42° (corresponding angles are equal in measure). y = 42.