Question

in the diagram below, \\( \angle dfg \cong \angle dgf \\) and \\( \angle edf \cong \angle gdh \\).
complete the proof that \\( \overline { d e } \cong \overline { d h } \\).
in the diagram, \\( \angle dfg \cong \angle dgf \\) and \\( \angle edf \cong \angle gdh \\). by the converse of the isosceles
triangle theorem, \\( \overline { d g } \cong \overline { d h } \\).
since \\( \angle efd \\) and \\( \angle dfg \\) form a linear pair, they are supplementary. so, for the same
reason, \\( \angle dgf \\) and \\( \angle dgh \\) are supplementary. by the congruent supplements theorem,
. so, \\( \triangle d e f \cong \triangle d h g \\) by the congruence
theorem. since congruent, \\( \overline { d e } \cong \overline { d h } \\).

Explanation:

Step1: Use Congruent Supplements Theorem

Since \(\angle DFG\cong\angle DGF\), \(\angle EFD\) and \(\angle DFG\) are supplementary (linear - pair), \(\angle DGH\) and \(\angle DGF\) are supplementary (linear - pair). By the Congruent Supplements Theorem, \(\angle EFD\cong\angle DGH\).

Step2: Determine triangle congruence

We know \(\angle EDF\cong\angle GDH\) (given), \(DF = DG\) (by the Converse of the Isosceles Triangle Theorem: if \(\angle DFG\cong\angle DGF\), then \(DF = DG\)), and \(\angle EFD\cong\angle DGH\) (from above). So, \(\triangle DEF\cong\triangle DHG\) by the Angle - Side - Angle (\(ASA\)) Congruence Theorem.

Answer:

\(\angle EFD\cong\angle DGH\); Angle - Side - Angle.