Question

given: \\( \overline { g h } \cong \overline { e h } \\), \\( \overline { d h } \cong \overline { f h } \\) and \\( \angle a d h \cong \angle b f h \\).
prove: \\( \overline { c d } \cong \overline { c f } \\).

Explanation:

Step1: Prove $\triangle D H G \cong \triangle F H E$

Since $\overline{G H} \cong \overline{E H}$, $\overline{D H} \cong \overline{F H}$, and $\angle D H G=\angle F H E$ (vertical angles are congruent), by the Side - Angle - Side (SAS) congruence criterion, $\triangle D H G \cong \triangle F H E$.

Step2: Get $D G = F E$

Because $\triangle D H G \cong \triangle F H E$, corresponding parts of congruent triangles are congruent (CPCTC), so $D G=F E$.

Step3: Prove $\angle C D G=\angle C F E$

Given $\angle A D H \cong \angle B F H$, then $180^{\circ}-\angle A D H = 180^{\circ}-\angle B F H$. Since $\angle C D G=180^{\circ}-\angle A D H$ and $\angle C F E = 180^{\circ}-\angle B F H$, we have $\angle C D G=\angle C F E$.

Step4: Prove $\triangle C D G \cong \triangle C F E$

We know that $\angle C=\angle C$ (common angle), $D G = F E$ (from Step 2), and $\angle C D G=\angle C F E$ (from Step 3). By the Angle - Side - Angle (ASA) congruence criterion, $\triangle C D G \cong \triangle C F E$.

Step5: Conclude $\overline{C D} \cong \overline{C F}$

Since $\triangle C D G \cong \triangle C F E$, by CPCTC, $\overline{C D} \cong \overline{C F}$.

Answer:

$\overline{C D} \cong \overline{C F}$ is proved as above.