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

If $\mathbf{AD}\mathbf{=}\mathbf{x}$and $\mathbf{BE}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{6}$then$\mathbf{x}\mathbf{=}\mathbf{?}$ 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 x for AD and $x+6$for $BE$into equation obtained in step 2.

$\begin{array}{c}x+6=2x\\ 2x-x=6\\ x=6\end{array}$

Therefore, the value of AD is 6 and BE is $6+6=12$.

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 12 for BE and 6 for AD into the equation obtained in step 3.

$\begin{array}{c}2BE=AD+CF\\ 2\times 12=6+CF\\ 24=6+CF\\ CF=18\end{array}$

Therefore, the values of x and CF are 6 and 18 respectively.