Question

let \overline{fg} be an altitude. which statement is true? angle dfg is congruent to angle efg. angle dgf is a right angle. dg = ge df = ef

Explanation:

Step1: Recall the definition of an altitude

An altitude of a triangle is a perpendicular segment from a vertex to the line containing the opposite side. So, if \(\overline{FG}\) is an altitude, then \(FG\perp ED\), which means \(\angle FGD = 90^{\circ}\) and \(\angle FGE=90^{\circ}\).

Step2: Analyze each option

• Option 1: \(\angle DFG\cong\angle EFG\)

There is no information given (such as \(FG\) being an angle - bisector) to suggest that these angles are congruent.

• Option 2: \(\angle DGF\) is a right angle

Since \(FG\) is an altitude (by definition, an altitude is perpendicular to the side it is drawn to), \(\angle DGF = 90^{\circ}\) (because \(FG\perp ED\)).

• Option 3: \(DG = GE\)

There is no information given (such as \(FG\) being a median) to suggest that \(G\) is the mid - point of \(ED\).

• Option 4: \(DF = EF\)

There is no information given (such as \(FG\) being a perpendicular bisector or the triangle being isosceles) to suggest that \(DF = EF\).

Answer:

\(\angle DGF\) is a right angle.