Question

For the following figure, given that $ABCD$is a trapezoid with median $\overline{MN}$.

Prove that $EF=\frac{1}{2}\left(AB-DC\right)$.

Short answer

Therefore, $EF=\frac{1}{2}\left(AB-DC\right)$.

Step-by-step solution

Step 1. Check the figure.

Consider the figure.

Step 2. Step description.

Since MN is a median therefore

$\begin{array}{l}ME=\frac{1}{2}DC\left(1\right)\\ MF=\frac{1}{2}AB\left(2\right)\end{array}$

Step 3. Step description.

From the equation (1) and (2) as follows:

$\begin{array}{l}EF=MF-ME\\ EF=\frac{1}{2}AB-\frac{1}{2}DC\\ EF=\frac{1}{2}\left(AB-DC\right)\end{array}$

Therefore, $EF=\frac{1}{2}\left(AB-DC\right)$.