Question

triangle fgh is formed by connecting the midpoints of the side of triangle cde. the lengths of the sides of triangle cde are shown. what is the length of \\(\overline{fg}\\)? figures not necessarily drawn to scale.

Explanation:

Step1: Recall the Midline Theorem

The Midline Theorem (also known as the Midsegment Theorem) states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half as long.

Step2: Identify the midline and the third side

In triangle \( CDE \), \( F \), \( G \), and \( H \) are midpoints. So, \( \overline{FG} \) connects the midpoints of \( CD \) and \( DE \), which means it should be parallel to \( CE \) and half its length. Wait, no, let's check the sides. Wait, looking at the diagram, \( EH = HC = 10/2 = 5 \)? Wait, no, the side \( CE \) is 10? Wait, no, the side \( CE \) is labeled 10? Wait, the diagram shows \( E \) to \( H \) to \( C \) with total length 10, and \( C \) to \( F \) to \( D \) with total length 8, and \( E \) to \( G \) to \( D \) with total length 8. Wait, actually, \( F \) is the midpoint of \( CD \), \( G \) is the midpoint of \( DE \), and \( H \) is the midpoint of \( CE \). Then, by the Midline Theorem, \( FG \) should be parallel to \( CE \) and \( FG=\frac{1}{2}CE \). Wait, but \( CE \) is 10? Wait, the length of \( CE \) is 10? Wait, the diagram shows the side \( CE \) (from \( E \) to \( C \)) with \( EH = HC \), so \( CE = 10 \). Then, \( FG \) is the midline, so \( FG=\frac{1}{2} \times CE \). Wait, no, maybe I got the sides wrong. Wait, let's re-examine: \( F \) is the midpoint of \( CD \), \( G \) is the midpoint of \( DE \), so the segment \( FG \) is the midline of triangle \( CDE \) parallel to \( CE \). So the length of \( FG \) should be half the length of \( CE \). The length of \( CE \) is 10? Wait, the diagram shows \( E \) to \( H \) to \( C \) with a total length of 10, so \( CE = 10 \). Therefore, \( FG=\frac{1}{2} \times 10 = 5 \)? Wait, no, wait, maybe I mixed up the sides. Wait, the side \( DE \) is 8 (from \( E \) to \( G \) to \( D \), so \( EG = GD = 4 \)), \( CD \) is 8 (from \( C \) to \( F \) to \( D \), so \( CF = FD = 4 \)), and \( CE \) is 10 (from \( E \) to \( H \) to \( C \), so \( EH = HC = 5 \)). Then, \( FG \) connects midpoints of \( CD \) and \( DE \), so it's parallel to \( CE \) and half its length. So \( FG = \frac{1}{2} \times CE = \frac{1}{2} \times 10 = 5 \)? Wait, but let's confirm the Midline Theorem: the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half as long. So if \( F \) is the midpoint of \( CD \) and \( G \) is the midpoint of \( DE \), then \( FG \parallel CE \) and \( FG = \frac{1}{2}CE \). Since \( CE = 10 \), then \( FG = 5 \). Wait, but let's check again. Wait, the problem says "Triangle FGH is formed by connecting the midpoints of the side of triangle CDE". So \( F \), \( G \), \( H \) are midpoints of \( CD \), \( DE \), \( CE \) respectively. Then, \( FG \) is the midline between \( CD \) and \( DE \), so parallel to \( CE \), length half of \( CE \). \( CE \) is 10, so \( FG = 5 \). Wait, but maybe I made a mistake. Wait, the side \( CE \) is 10? The diagram shows \( E \) to \( H \) to \( C \) with a label of 10, so \( CE = 10 \). Therefore, \( FG = \frac{1}{2} \times 10 = 5 \).

Wait, no, wait, maybe the side is \( DE \) or \( CD \). Wait, no, let's look at the labels: \( E \) to \( D \) is 8 ( \( EG = GD = 4 \) ), \( C \) to \( D \) is 8 ( \( CF = FD = 4 \) ), \( C \) to \( E \) is 10 ( \( EH = HC = 5 \) ). Then, \( FG \) connects midpoints of \( CD \) ( \( F \) ) and \( DE \) ( \( G \) ), so by Midline Theorem, \( FG \parallel CE \) and \( FG = \frac{1}{2} CE \). So \( CE = 10 \), so \( FG = 5 \).

Step1: Apply Midline Theorem

The Midline Theorem states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half its length. In \( \triangle CDE \), \( F \) (midpoint of \( CD \)) and \( G \) (midpoint of \( DE \)) form \( \overline{FG} \), which is parallel to \( \overline{CE} \) and \( FG = \frac{1}{2} CE \).

Step2: Determine length of \( CE \)

From the diagram, \( CE = 10 \) (since \( EH + HC = 10 \) and \( H \) is the midpoint, so \( CE = 10 \)).

Step3: Calculate \( FG \)

Using the Midline Theorem: \( FG = \frac{1}{2} \times CE = \frac{1}{2} \times 10 = 5 \).

Answer:

\( \boxed{5} \)