Question

in the diagram below, \\( \overline { e f } \perp \overline { f g } \\) and \\( \overline { f h } \perp \overline { e g } \\).

complete the proof that \\( \frac { e f } { e h } = \frac { e g } { e f } \\).

since \\( \overline { e f } \perp \overline { f g } \\) and \\( \overline { f h } \perp \overline { e g } \\), \\( \angle e f g \\) and \\( \angle e h f \\) are

so, because all right angles are

congruent,

Explanation:

Step1: Identify right angles

Since \( \overline{EF}\perp\overline{FG}\) and \( \overline{FH}\perp\overline{EG}\), \( \angle EFG = 90^{\circ}\) and \( \angle EHF=90^{\circ}\). So, \( \angle EFG\) and \( \angle EHF\) are right angles.

Step2: Use AA (Angle - Angle) similarity criterion

In \( \triangle EHF\) and \( \triangle EFG\), \( \angle E=\angle E\) (common angle) and \( \angle EHF=\angle EFG = 90^{\circ}\). By AA similarity criterion, \( \triangle EHF\sim\triangle EFG\).

Step3: Apply the property of similar triangles

If two triangles \( \triangle ABC\) and \( \triangle DEF\) are similar (\( \triangle ABC\sim\triangle DEF\)), then the ratios of their corresponding sides are equal. For \( \triangle EHF\sim\triangle EFG\), we have \(\frac{EF}{EH}=\frac{EG}{EF}\) (corresponding sides of similar triangles).

Answer:

\(\frac{EF}{EH}=\frac{EG}{EF}\) is proved by showing \( \triangle EHF\sim\triangle EFG\) using the AA similarity criterion.