Question

given: △peg, $overline{ep}congoverline{eg}$, altitudes $overline{pk}$ and $overline{gh}$
prove: $overline{pk}congoverline{gh}$

1. △peg, $overline{ep}congoverline{eg}$, altitudes $overline{pk}$ and $overline{gh}$
2. given
3. ∠phg and ∠gkp are right angles.

2.
3.

1. all right angles are congruent.
2. ∠hpg≅∠kgp

4.

Explanation:

Step1: State the given information

Given $\triangle PEG$, $\overline{EP}\cong\overline{EG}$, altitudes $\overline{PK}$ and $\overline{GH}$

Step2: Define the nature of angles formed by altitudes

By the definition of an altitude, $\angle PHG$ and $\angle GKP$ are right - angles.

Step3: Apply the property of right angles

All right angles are congruent, so $\angle PHG\cong\angle GKP$.

Step4: Use the property of an isosceles triangle

In $\triangle PEG$, since $\overline{EP}\cong\overline{EG}$, the base angles $\angle EPG$ and $\angle EGP$ are congruent. And $\angle HPG$ and $\angle EPG$, $\angle KGP$ and $\angle EGP$ are related such that $\angle HPG\cong\angle KGP$. Then, by the Angle - Angle (AA) similarity criterion for right - triangles $\triangle PHG\cong\triangle GKP$ (by AA similarity, as we have two pairs of congruent angles), and corresponding parts of congruent triangles are congruent, so $\overline{PK}\cong\overline{GH}$.

Answer:

1. Given
2. Definition of altitude (An altitude of a triangle is a perpendicular segment from a vertex to the opposite - side, so the angles formed are right - angles)
3. All right angles are congruent
4. Base angles of an isosceles triangle are congruent (Since $\overline{EP}\cong\overline{EG}$ in $\triangle PEG$, $\angle EPG=\angle EGP$, and $\angle HPG$ and $\angle EPG$, $\angle KGP$ and $\angle EGP$ are the same or related in such a way that $\angle HPG\cong\angle KGP$)