Question

given: △peg, (overline{ep}congoverline{eg}), altitudes (overline{pk}) and (overline{gh}). prove: (overline{pk}congoverline{gh}). 5. click here to insert. 6. △phg≅△gkp. 7. click here to insert.

Explanation:

Step1: Define right - angles

Since \(\overline{PK}\) and \(\overline{GH}\) are altitudes of \(\triangle PEG\), \(\angle PKE = \angle GHE=90^{\circ}\)

Step2: Identify base - angles of isosceles triangle

Given \(\overline{EP}\cong\overline{EG}\), in \(\triangle PEG\), \(\angle EPG=\angle EGP\)

Step3: Prove \(\triangle PHG\cong\triangle GKP\)

We have two right - angles (\(\angle PHG=\angle GKP\)), equal non - right angles (\(\angle HPG=\angle KGP\)) and a common side \(\overline{PG}\), so by AAS \(\triangle PHG\cong\triangle GKP\)

Step4: Find congruent segments

As \(\triangle PHG\cong\triangle GKP\), \(\overline{PK}\cong\overline{GH}\) (corresponding parts of congruent triangles are congruent)

Answer:

1. Step - 1: Identify right - angled triangles

In \(\triangle PEG\), since \(\overline{PK}\) and \(\overline{GH}\) are altitudes, \(\angle PKE=\angle GHE = 90^{\circ}\)

1. Step - 2: Use the property of isosceles triangle

Given \(\overline{EP}\cong\overline{EG}\), then \(\angle EPG=\angle EGP\) (base - angles of an isosceles triangle are equal)

1. Step - 3: Prove triangle congruence

In \(\triangle PHG\) and \(\triangle GKP\):

• \(\angle PHG=\angle GKP = 90^{\circ}\) (altitudes)
• \(\angle HPG=\angle KGP\) (from step 2)
• \(\overline{PG}=\overline{GP}\) (common side)
• So, \(\triangle PHG\cong\triangle GKP\) (by AAS congruence criterion)

1. Step - 4: Use the property of congruent triangles

Since \(\triangle PHG\cong\triangle GKP\), corresponding parts of congruent triangles are congruent. So, \(\overline{PK}\cong\overline{GH}\)