In Exercises 12-20, $\mathbf{TA}\mathbf{=}\mathbf{AB}\mathbf{=}\mathbf{BC}$and $\mathbf{TD}\mathbf{=}\mathbf{DE}\mathbf{=}\mathbf{EF}$

If$\mathbf{AD}\mathbf{=}\mathbf{7}$ then$\mathbf{BE}\mathbf{=}\mathbf{?}$ and $\mathbf{CF}\mathbf{=}\mathbf{?}$.

Consider $\Delta TBE$. From the figure it can be observed that $TA=AB\text{and}TD=DE$. AD is the midsegment of $\Delta TBE$. The midsegment theorem states that the length of the midsegment is half of the base of the triangle. Therefore,

$\begin{array}{c}AD=\frac{BE}{2}\\ BE=2AD\end{array}$

Substitute 7 for AD into the equation obtained in step 2.

$\begin{array}{c}BE=2\times 7\\ =14\end{array}$

Consider a trapezoid CFDA. From the figure, it can be observed that $BC=AB\text{and}EF=DE$. BE is the median of trapezoid CFDA. According to the property of the median of a trapezoid, the length of the median is the average of the two base lengths. Therefore,

$\begin{array}{c}BE=\frac{1}{2}\left(AD+CF\right)\\ 2BE=AD+CF\end{array}$

Substitute 14 for BE and 7 for AD into the equation obtained in step 3.

$\begin{array}{c}2BE=AD+CF\\ 2\times 14=7+CF\\ 28=7+CF\\ CF=21\end{array}$

Therefore, the values of BE and CF are 14 and 21 respectively.