Question

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Explanation:

Problem 6:

Step1: Recall Segment Addition

For segment \( \overline{EFGH} \), we have \( EG = EF + FG \) and \( FH = FG + GH \) (Segment Addition Postulate).

Step2: Use Given Equality

Given \( EG = FH \), substitute the expressions from Step 1: \( EF + FG = FG + GH \).

Step3: Subtract \( FG \) from Both Sides

Subtract \( FG \) from each side of the equation: \( EF + FG - FG = FG + GH - FG \), which simplifies to \( EF = GH \).

Step1: Midpoint Definition

Since \( C \) is the midpoint of \( \overline{BD} \), \( BC = CD \) (Definition of Midpoint).

Step2: Add \( AB \) and \( DE \)

Given \( AB = DE \). Add \( AB \) to \( BC \) and \( DE \) to \( CD \): \( AB + BC = DE + CD \).

Step3: Segment Addition

By Segment Addition, \( AB + BC = AC \) and \( DE + CD = CE \). So \( AC = CE \).

Step4: Midpoint Definition

If \( AC = CE \), then \( C \) is the midpoint of \( \overline{AE} \) (Definition of Midpoint).

Step1: Definition of Complementary Angles

Since \( \angle 1 \) and \( \angle 2 \) are complementary, \( m\angle 1 + m\angle 2 = 90^\circ \) (Definition of Complementary Angles). Similarly, \( m\angle 3 + m\angle 4 = 90^\circ \) because \( \angle 3 \) and \( \angle 4 \) are complementary.

Step2: Substitute \( m\angle 2 = m\angle 4 \)

Given \( m\angle 2 = m\angle 4 \). From Step 1, \( m\angle 1 = 90^\circ - m\angle 2 \) and \( m\angle 3 = 90^\circ - m\angle 4 \). Substitute \( m\angle 2 \) with \( m\angle 4 \) in the expression for \( m\angle 1 \): \( m\angle 1 = 90^\circ - m\angle 4 \).

Step3: Conclude \( m\angle 1 = m\angle 3 \)

Since \( m\angle 3 = 90^\circ - m\angle 4 \) and \( m\angle 1 = 90^\circ - m\angle 4 \), we have \( m\angle 1 = m\angle 3 \).

Answer:

\( EF = GH \) is proven.

Problem 7: