Question

in the figure below, points d, e, and f are the midpoints of the sides of △abc. suppose ab = 48, bc = 56, and de = 32. find the following lengths. figure of triangle abc with midpoints d, e, f df = ac = cf =

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: Find \( DF \)

Points \( D \) and \( F \) are midpoints of \( AB \) and \( AC \) respectively? Wait, no, let's look at the triangle. Wait, \( D \) is midpoint of \( AB \), \( E \) midpoint of \( BC \), \( F \) midpoint of \( AC \). Then \( DE \) is midline parallel to \( AC \), \( DF \) is midline parallel to \( BC \), and \( EF \) is midline parallel to \( AB \).

So, the midline \( DF \) connects midpoints of \( AB \) and \( AC \), so it should be parallel to \( BC \) and half its length. Wait, \( BC = 56 \), so \( DF=\frac{1}{2}BC \). Let's check: \( BC = 56 \), so \( DF=\frac{56}{2}=28 \).

Step3: Find \( AC \)

The midline \( DE \) connects midpoints of \( AB \) and \( BC \), so it should be parallel to \( AC \) and half its length. So \( DE=\frac{1}{2}AC \). Given \( DE = 32 \), then \( AC = 2\times DE = 2\times32 = 64 \).

Step4: Find \( CF \)

Since \( F \) is the midpoint of \( AC \), \( CF=\frac{1}{2}AC \). We found \( AC = 64 \), so \( CF=\frac{64}{2}=32 \).

Answer:

$DF = 28$, $AC = 64$, $CF = 32$