Given:
AG = 8 inches
BF is the mid-segment of triangle ADG.
CE is the mid-segment of triangle DBF.
Solution.
A midsegment of a triangle is a segment that connects the midpoints of two sides of a triangle.
From the mid-segment theorem, the segment joining the midpoints of two sides of a triangle is parallel to and half the length of the third side.
Hence,
[tex]\begin{gathered} BF\text{ = }\frac{1}{2}\text{ }\times\text{ AG } \\ =\text{ }\frac{1}{2}\text{ }\times\text{ 8} \\ =\text{ 4 inches} \end{gathered}[/tex]
Similarly, CE is a mid-segment of the triangle DBF, hence:
[tex]\begin{gathered} CE\text{ = }\frac{1}{2}\text{ }\times\text{ BF } \\ =\text{ }\frac{1}{2}\text{ }\times\text{ 4} \\ =\text{ 2 inches} \end{gathered}[/tex]
Answer: 2 inches (option A)
