Chapter 7.5, Problem 22WE

To determine

To Prove: For the given condition, $\overrightarrow{MN}\parallel \overrightarrow{BC};MN=\frac{1}{2}BC$.

Explanation of Solution

Given information:

M is the midpoint of AB.

N is the midpoint of AC.

Extend the line MN to point E such that MN= NE and join CE.

The figure is shown below.

  

In $△AMN$ and $\text{Δ}CEN$ , note that

  $\angle ANM=\angle CNE$ (Vertically Opposite angles)

  $AN=NC(N$ is midpoint of $AC)$

  $MN=EN$ (By construction)

This implies that $△AMN\cong \text{Δ}CEN$ ( by SAS postulate)

Since corresponding parts of congruent triangles are congruent.

This implies that $AM=CE$ , but as $M$ is the midpoint of AB, then AM= BM

So it follows that

BM= CE also note that $\angle MAN=\angle ECN$ (since $\text{Δ}AMN\cong \text{Δ}CEN)$ )

But these are alternate interior angles formed by the lines CE and AM and by transversal CA.

Therefore, $CE\parallel AM$ which further implies that $CE\parallel BM$

Since one pair of opposite sides of quadrilateral BCEM are parallel and equal.

Therefore BCEM is parallelogram so that

  $ME\parallel BC$ And $ME=BC$

Since $MN=\frac{1}{2}ME,$ , this implies that $MN\parallel BC$ and $MN=\frac{1}{2}BC$ .