Question

use the image to answer the question.
given the image, which statement must be true?
(1 point)
\\(\bigcirc\\ de = \frac{1}{2}ab\\)
\\(\bigcirc\\ df = \frac{1}{2}bc\\)
\\(\bigcirc\\ 2df = ac\\)
\\(\bigcirc\\ de = df\\)

Explanation:

Step1: Identify Midsegments

From the diagram, \( D \) is the midpoint of \( AB \) (since \( AD = DB \)) and \( E \) is the midpoint of \( BC \) (since \( BE = EC \)), \( F \) is the midpoint of \( AC \) (since \( AF = FC \)). By the Midsegment Theorem (or Midline Theorem) in triangles, the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half its length.

Step2: Analyze Each Option

• For \( DE \): \( D \) (midpoint of \( AB \)) and \( E \) (midpoint of \( BC \)) form \( DE \). By Midsegment Theorem, \( DE=\frac{1}{2}AC \), not \( \frac{1}{2}AB \). So first option is false.
• For \( DF \): \( D \) (midpoint of \( AB \)) and \( F \) (midpoint of \( AC \)) form \( DF \). By Midsegment Theorem, \( DF = \frac{1}{2}BC \). This matches the second option.
• For \( 2DF = AC \): From Midsegment Theorem, \( DF=\frac{1}{2}BC \), so \( 2DF = BC \), not \( AC \). Third option is false.
• For \( DE = DF \): \( DE=\frac{1}{2}AC \) and \( DF=\frac{1}{2}BC \). There's no indication \( AC = BC \), so \( DE \) and \( DF \) aren't necessarily equal. Fourth option is false.

Answer:

\( DF = \frac{1}{2}BC \) (the second option, i.e., the option with \( DF = \frac{1}{2}BC \))