Exercise 133 Page 328

Practice makes perfect

a Let's start identifying the angles! If we view AD as a transversal, we can identify two pairs of corresponding angles with respect to the lines AB and EC, which intersect the transversal. Since AB ∥ EC, we know they are corresponding by the Corresponding Angles Theorem.

Having found these angles, we can identify two pairs of same-side interior angles — one pair with respect to CE and another with respect to BE. Since BC ∥ AD, we know that these are congruent by the Same-Side Interior Angles Theorem.

For each of the three triangles we can identify in the trapezoid, we know two angles. With this information by the Triangle Angle Sum Theorem we can identify the third angle. & △ ABE & m∠ B+60^(∘)+40^(∘)=180^(∘) ⇔ m∠ B=80^(∘) [1em] & △ BEC & m∠ E+60^(∘)+40^(∘)=180^(∘) ⇔ m∠ E=80^(∘) [1em] & △ CDE & m∠ C+60^(∘)+40^(∘)=180^(∘) ⇔ m∠ C=80^(∘) Let's complete the diagram with these angles.

Finally, we will summarize the angle measures. m∠ A = 40^(∘) m∠ ABE = 80^(∘) m∠ EBC = 60^(∘) m∠ BCE = 40^(∘) m∠ ECD = 40^(∘) m∠ D = 60^(∘) m∠ DEC = 40^(∘) m∠ CEB = 80^(∘) m∠ BEA = 60^(∘)

b In Part A we identified all angles in the three different triangles we can see inside of the trapezoid. We already know m∠ A and m∠ D, so what remains to be found is m∠ ABC and m∠ BCD.

m∠ ABC:& m∠ ABE+m∠ EBC m∠ BCD:& m∠ BCE+m∠ ECDBy substituting the values of m∠ ABE, m∠ EBC, m∠ BCE, and m∠ ECD, we can find these angle measures. m∠ ABC:& 80^(∘)+60^(∘) = 140^(∘) m∠ BCD:& 40^(∘)+80^(∘) = 120^(∘) Now we have enough information to determine the sum of the angles that make up the trapezoid ABCD.

m∠ A+m∠ ABC+m∠ BCD+m∠ D

SubstituteValues

Substitute values

40^(∘)+140^(∘)+120^(∘)+60^(∘)

AddTerms

Add terms

360^(∘)

The angles sum to 360^(∘).