91Ó°ÊÓ

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

$MNâˆ\yen BC$

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

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

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

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

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

$\begin{array}{c}\stackrel{¯}{AB}=\frac{1}{2}\stackrel{¯}{CD}\\ =\frac{1}{2}×24\\ =12\end{array}$

and

$\begin{array}{c}\stackrel{¯}{EF}=\frac{1}{2}\stackrel{¯}{CD}\\ =\frac{1}{2}×24\\ =12\end{array}$

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