Answered

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Given: [tex]$\triangle ABC$[/tex]

Prove: A midsegment of [tex]$\triangle ABC$[/tex] is parallel to a side of [tex]$\triangle ABC$[/tex].

[tex]\[
\begin{array}{|c|c|}
\hline
\text{Statement} & \text{Reason} \\
\hline
1. \text{Define the vertices of } \triangle ABC \text{ to have unique points } A(x_1, y_1), B(x_2, y_2), \text{ and } C(x_3, y_3) & \text{Given} \\
\hline
2. \text{Let } D \text{ be the midpoint of } \overline{AB} \text{ and } E \text{ be the midpoint of } \overline{BC} & \text{Defining midpoints} \\
\hline
3. D=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right), \quad E=\left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}\right) & \text{Definition of midpoint} \\
\hline
\end{array}
\][/tex]

What is the reason for statement 3 in this proof?

A. using point-slope formula
B. definition of parallel lines
C. Transitive Property of Equality
D. Reflexive Property of Equality
E. definition of midpoint

Sagot :

GinnyAnswer
The reason for statement 3 in this proof is:

E. definition of midpoint

Statement 3 gives the coordinates of points [tex]\(D\)[/tex] and [tex]\(E\)[/tex], which are midpoints of line segments [tex]\(\overline{AB}\)[/tex] and [tex]\(\overline{BC}\)[/tex], respectively. According to the definition of a midpoint, the coordinates of a midpoint are the averages of the coordinates of the endpoints of the line segment.

- For point [tex]\(D\)[/tex], which is the midpoint of [tex]\(\overline{AB}\)[/tex], the coordinates are calculated as [tex]\(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)[/tex].
- For point [tex]\(E\)[/tex], which is the midpoint of [tex]\(\overline{BC}\)[/tex], the coordinates are calculated as [tex]\(\left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}\right)\)[/tex].

Therefore, the coordinates given in statement 3 align with the definition of a midpoint.
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