Points A, B, E and F are midpoints of $\overline{XC}$, $\overline{XD}$, $\overline{YC}$and $\overline{YD}$. Complete.

If $\text{CD}=\mathbf{24}$, then $AB=?$and $EF=?$ .

It is given that two triangles XCD and YCD having the same base with A, B, E and F are the midpoints of sides $\overline{XC}$, $\overline{XD}$, $\overline{YC}$and $\overline{YD}$ respectively.

The following theorems are used.

1) Theorem 5-11 - In a triangle the line joining the midpoints of two legs will be parallel to the third leg and will be equal to half its measure.

In the triangle ABC given below, M and N are midpoints of sides AB and AC respectively. Therefore, by the theorem,

$MN\parallel BC$

$MN=\frac{1}{2}BC$

Here, A, B, E and F are the midpoints of sides $\overline{XC}$, $\overline{XD}$, $\overline{YC}$ and $\overline{YD}$ respectively.

Using the theorem presented above, in triangles XCD and YCD the sides $\overline{AB}$ and $\overline{EF}$can be expressed as,

$\begin{array}{l}\overline{AB}=\frac{1}{2}\overline{CD}\\ \overline{EF}=\frac{1}{2}\overline{CD}\end{array}$

It is given that $\overline{CD}=24$. Substituting this in the above equation, the measures of the sides $\overline{AB}$and $\overline{EF}$can be obtained as,

$\begin{array}{c}\overline{AB}=\frac{1}{2}\overline{CD}\\ =\frac{1}{2}\times 24\\ =12\end{array}$

and

$\begin{array}{c}\overline{EF}=\frac{1}{2}\overline{CD}\\ =\frac{1}{2}\times 24\\ =12\end{array}$

Therefore, $\overline{AB}=12$ and $\overline{EF}=12$.