Exercise 18 Page 370

By the Reflexive Property of Congruent Segments we get GH ≅ GH. Below we list the corresponding congruent parts of both triangles. cc ∠ F ≅ ∠ J & Angle ∠ GHF ≅ ∠ HGJ & Angle GH ≅ GH & Non-included Side Therefore, by applying the Angle-Angle-Side (AAS) Congruence Postulate we conclude that △ FHG ≅ △ JGH. Consequently, by definition of congruent polygons, we get FH ≅ JG.

Paragraph Proof

Given: & ∠ F ≅ ∠ J, FH ∥ GJ Prove: & FH ≅ JG Proof: Since FH ∥ GJ and GH is a transversal, by the Alternate Interior Angles Theorem we obtain that ∠ GHF ≅ ∠ HGJ. Also, by applying the Reflexive Property of Congruent Segments we get GH ≅ GH.

Additionally, we are told that ∠ F ≅ ∠ J and so, by the Angle-Angle-Side (AAS) Congruence Postulate we conclude that △ FHG ≅ △ JGH, which implies that FH ≅ JG.