Question

For the following figure, given that $TA=TB=BC$and $TD=DE=EF$.

Mike makes up a problem for the figure, setting $AD=2x+1$, $BE=4x+2$and $CF=6x-3$ and asking for the value of x. Katie says, “Anybody can do that.” Explain

Short answer

This figure is possible for all values of x.

Step-by-step solution

Step 1. Check the figure.

Consider the figure.

Step 2. Step description.

Consider that $TA=AB=BC,TD=DE=EF,AD=2x+1,BE=4x+2,CF=6x+3$.

Use the mid-segment or median of the trapezoid as follows:

$\begin{array}{l}AD=\frac{BE}{2}\\ AD=\frac{4x+2}{2}\\ AD=2x+1\end{array}$

Step 3. Step description.

Similarly,

$\begin{array}{l}BE=\frac{AD+CF}{2}\\ BE=\frac{2x+1+\left(6x+3\right)}{2}\\ BE=\frac{8x+4}{2}\\ BE=4x+2\end{array}$

Therefore, the figure is possible for all values of x.