Let's copy the diagram from the book indicating the parallel bases and congruent legs of the isosceles trapezoid. To show that each pair of base angles are congruent, let's draw a line through H that is parallel to GF, and extend this line until it meets the line through F and J.
We will first show that △ JHN is an isosceles triangle, then we will compare the base angles of the isosceles trapezoid to the congruent base angles of this triangle. We do this in several steps, focusing on different parts of the diagram in different steps. Let's start with looking at quadrilateral GHNF.
By construction, the opposite sides of GHNF are parallel, so this quadrilateral is a parallelogram. According to Theorem 6.3, this means that opposite sides are congruent.
GF≅HN
Let's turn our attention now to triangle △ JHN.
We just proved that GF is congruent to HN. Since FGHJ is an isosceles trapezoid, GF is also congruent to HJ. By the Transitive Property of Congruence, this means that two sides of triangle △ JHN are congruent.
HJ≅HN
Since △ JHN is an isosceles triangle, its base angles are congruent.
∠ HJN≅ ∠ HNJ
Let's compare these base angles to the base angles of FGHJ.
Angles at G and H
Quadrilateral GHNF is a parallelogram, so according to Theorem 6.4, opposite angles are congruent.
∠ G≅ ∠ HNJ
Segment HJ is a transversal to the parallel segments GH and FN, so according to the Alternate Interior Angles Theorem, the angles at H and J are congruent.
∠ HJN≅ ∠ H
Let' put these observations together.
∠ G≅ ∠ HNJ≅ ∠ HJN≅ ∠ H
The Transitive Property of Congruence guarantees that ∠ G and ∠ H are congruent.
Angles at F and J
Let's concentrate now on ∠ F and ∠ J.
Angles ∠ F and ∠ HNJ are consecutive angles of parallelogram GHNF, so according to Theorem 6.5, these are supplementary angles.
Angles ∠ J and ∠ HJN form a linear pair, so these are also supplementary.
Since ∠ F and ∠ J are supplementary to the congruent base angles of the isosceles triangle △ JHN, these are congruent.
Let's summarize the steps above in a paragraph as asked in the question.
Completed Proof
2
&Given:&& FGHJ is an isosceles trapezoid with
& && GH∥FJandFG≅JH
&Prove:&& ∠ G≅∠ H and∠ F≅∠ J
Proof:
Let's add triangle △ JHN to trapezoid FGHJ to form parallelogram GHNF. Opposite sides of a parallelogram are congruent, so GF is congruent to HN. It is given that GF is congruent to HJ, so by the transitive property of congruence, HN is congruent to HJ. According to the Isosceles Triangle Theorem, this congruence implies that in triangle △ JHN, angles ∠ HJN and ∠ HNJ are congruent. Angle ∠ HNJ is congruent to ∠ G (because these are opposite angles of parallelogram GHNF) and angle ∠ HJN is congruent to ∠ H (because these are alternate interior angles between parallel lines), so by transitivity, ∠ G≅∠ H. Angle ∠ HNJ is supplementary to ∠ F (because these are consecutive angles of parallelogram GHNF) and angle ∠ HJN is supplementary to ∠ J (because these form a linear pair), so ∠ F≅∠ J (because they are supplements of congruent angles).
