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

If $\mathbf{AD}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{y}$,$\mathbf{BE}\mathbf{=}\mathbf{20}$ and $\mathbf{CF}\mathbf{=}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{y}$then $\mathbf{CF}\mathbf{=}\mathbf{?}$(numerical answer).

Consider $\Delta TBE$. From the figure it can be observed that $TA=AB\text{and}TD=DE$. 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 $x+y$for AD and 20 for BE into equation obtained in step 2.

$\begin{array}{c}20=2\left(x+y\right)\\ x+y=\frac{20}{2}\\ y=10-x\end{array}$

Consider a trapezoid CFDA. From the figure, it can be observed that $BC=AB\text{and}EF=DE$. 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 20 for BE, $x+y$ for AD and $4x-y$ for CF into equation obtained in step 3.

$\begin{array}{c}2BE=AD+CF\\ 2\times 20=x+y+4x-y\\ 40=5x\\ x=\frac{40}{5}\\ =8\end{array}$

Substitute 8 for x into $y=10-x$ to find the value of y.

$\begin{array}{c}y=10-x\\ =10-8\\ =2\end{array}$

Substitute 8 for x and 2 for y into $CF=4x-y$ to find the value of CF.

$\begin{array}{c}CF=4\times 8-2\\ =32-2\\ =30\end{array}$

Therefore, the value of CF is 30.