Question

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

Write an equation that relates $\mathrm{AD},\mathrm{BE}$and $\mathrm{CF}$. (Hint: Think of trapezoid CFDA).

Short answer

The equation that relates AD, BEand CF is $\mathbf{2}\mathbf{BE}\mathbf{=}\mathbf{AD}\mathbf{+}\mathbf{CF}$.

Step-by-step solution

Step 1. Highlight a trapezoid in the given figure.

Step 2. Description of step.

Consider a trapezoid CFDA. From the figure, it can be observed that $BC=AB\text{and}EF=DE$.

Step 3. Description of step.

From the figure, it can be observed that 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,

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

Therefore, the equation that relates AD, BEand CF is $2BE=AD+CF$.