Question

1. choose the correct answer.

$overline{df}$, $overline{ef}$, and $overline{de}$ are midsegments of $\triangle abc$
$be = 3$ cm
$af = 2$ cm
what is the length of $overline{df}$?

• 4 cm
• 2 cm
• 6 cm
• 3 cm

Explanation:

Step1: Recall Midsegment Theorem

The midsegment of a triangle is parallel to the third side and half its length. Also, midsegments connect midpoints of sides. So, \( E \) is midpoint of \( BC \), \( F \) midpoint of \( AC \), \( D \) midpoint of \( AB \).

Step2: Identify \( DF \) relation to \( BC \)

\( DF \) is a midsegment, so \( DF \parallel BC \) and \( DF=\frac{1}{2}BC \). But also, since \( E \) is midpoint, \( BE = EC \). Wait, \( BE = 3 \) cm, so \( BC = BE + EC = 3 + 3 = 6 \) cm? No, wait, \( DF \) is midsegment, but also, maybe \( DF \) is equal to \( BE \)? Wait, no, let's check again. Wait, midsegment: \( DF \) connects midpoints of \( AB \) and \( AC \), so \( DF \parallel BC \) and \( DF=\frac{1}{2}BC \). But \( BE = 3 \) cm, and \( E \) is midpoint, so \( BC = 2BE = 6 \) cm? No, wait, no, \( DF \) is midsegment, but maybe I made a mistake. Wait, no, the midsegment theorem: the segment connecting midpoints of two sides is parallel to the third side and half its length. So \( D \) is midpoint of \( AB \), \( F \) is midpoint of \( AC \), so \( DF \parallel BC \) and \( DF=\frac{1}{2}BC \). But \( BE = 3 \) cm, and \( E \) is midpoint of \( BC \), so \( BC = 2BE = 6 \) cm? No, that would make \( DF = 3 \) cm? Wait, no, wait the options: 4, 2, 6, 3. Wait, maybe \( DF \) is equal to \( BE \)? Wait, \( E \) is midpoint, \( D \) is midpoint, so \( DE \parallel AC \), \( EF \parallel AB \), so \( ADFE \) is a parallelogram? Wait, \( D \) midpoint of \( AB \), \( F \) midpoint of \( AC \), \( E \) midpoint of \( BC \). So \( DF \parallel BC \), \( DE \parallel AC \), \( EF \parallel AB \). So \( DF = \frac{1}{2}BC \), \( DE = \frac{1}{2}AC \), \( EF = \frac{1}{2}AB \). But \( BE = 3 \) cm, and \( BE \) is half of \( BC \)? No, \( BE = 3 \) cm, so \( BC = 6 \) cm, so \( DF = 3 \) cm? Wait, but \( BE = 3 \) cm, and \( DF \) is midsegment, so \( DF = BE \)? Wait, \( BE = 3 \) cm, so \( DF = 3 \) cm? Wait the options have 3 cm as an option. Wait, let's check again. \( BE = 3 \) cm, \( E \) is midpoint, so \( BC = 6 \) cm. Then \( DF = \frac{1}{2}BC = 3 \) cm. Yes, that makes sense. So \( DF = 3 \) cm.

Step3: Match with options

The option with 3 cm is the last one. So the length of \( DF \) is 3 cm.

Answer:

3 cm (the option with 3 cm, i.e., the last option: 3 cm)