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

If $\mathbf{AD}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{3}$, $\mathbf{BE}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{y}$and $\mathbf{CF}\mathbf{=}\mathbf{36}$then$\mathbf{x}\mathbf{=}\mathbf{?}$ and $\mathbf{y}\mathbf{=}\mathbf{?}$.

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 $x+3$ for AD and $x+y$ for BE into equation obtained in step 2.

$\begin{array}{c}x+y=2\left(x+3\right)\\ x+y=2x+6\\ x=y-6\end{array}$

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 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 $x+y$ for BE, $x+3$ for AD and 36 for CF into equation obtained in step 3.

$\begin{array}{c}2BE=AD+CF\\ 2\left(x+y\right)=x+3+36\\ 2x+2y=x+39\\ x+2y=39\end{array}$

Substitute $y-6$ for x into equation obtained in step 4.

$\begin{array}{c}y-6+2y=39\\ 3y=45\\ y=\frac{45}{3}\\ =15\end{array}$

Substitute 15 for y into $x=y-6$to find the value of x.

$\begin{array}{c}x=15-6\\ =9\end{array}$

Therefore, the values of x and y are 9 and 15 respectively.