Question

Given: $▱\text{ABCD}$; $\overline{\text{BE}}\parallel \overline{\text{MD}}$;

M is the midpoint of $\overline{\text{AB}}$;

Prove: $\text{DE}=\text{BC}$

Short answer

It is proved $\overline{DE}=\overline{BC}$.

Step-by-step solution

Step 1. Consider the diagram.

Here ABCD is a parallelogram, $\overline{BE}\parallel \overline{MD}$and M is the midpoint of $\overline{AB}$.

Step 2. Show the prove.

In $\Delta ABE$, $\overline{MD}\parallel \overline{BE}$ and M is the midpoint of $\overline{AB}$.

So,

D is a midpoint of $\overline{AE}$.

(According to the converse of midpoint theorem, if a line is drawn through the midpoint of a side of a triangle parallel to the other side, then it bisects the third side.)

As D is a midpoint of $\overline{AE}$,

$\overline{AD}=\overline{DE}$ …… (i)

Now,

ABCD is a parallelogram.

So,

$\overline{AD}=\overline{BC}$ …… (ii) (Opposite sides of a parallelogram are equal.)

From (i) and (ii),

$\overline{DE}=\overline{BC}$

Step 3. State the conclusion.

Therefore, $\overline{DE}=\overline{BC}$(proved).