Question

in exercises 15 and 16, use the information given in the diagram to write a proof. 15. prove \\( \overline{fl} \cong \overline{hn} \\)

Explanation:

Step1: Identify Given Congruences

From the diagram, \( FG \cong HG \), \( JG \cong KG \), \( JM \cong KM \), \( LM \cong NM \). Also, \( \angle FGJ \cong \angle HGK \) (vertical angles), \( \angle JML \cong \angle KMN \) (vertical angles).

Step2: Prove \( \triangle FGJ \cong \triangle HGK \)

By SAS (Side - Angle - Side): \( FG = HG \), \( \angle FGJ=\angle HGK \), \( JG = KG \). So \( \triangle FGJ\cong\triangle HGK \), thus \( FJ = HK \).

Step3: Prove \( \triangle JML \cong \triangle KMN \)

By SAS: \( JM = KM \), \( \angle JML=\angle KMN \), \( LM = NM \). So \( \triangle JML\cong\triangle KMN \), thus \( JL = KN \).

Step4: Prove \( FL\cong HN \)

\( FL=FJ + JL \), \( HN = HK+KN \). Since \( FJ = HK \) (from Step 2) and \( JL = KN \) (from Step 3), by the Segment Addition Postulate and Substitution, \( FL = HN \), so \( \overline{FL}\cong\overline{HN} \).

Answer:

By proving \( \triangle FGJ \cong \triangle HGK \) (SAS) and \( \triangle JML \cong \triangle KMN \) (SAS), then using segment addition and substitution, we show \( FL = HN \), so \( \overline{FL}\cong\overline{HN} \).