Recall the Triangle Midsegment Theorem:
This implies that since BE is a midsegment of the triangle (as B and E are midpoints), then:
[tex]BE=\frac{1}{2}CD[/tex]
Substitute BE=5 into the equation and solve for CD:
[tex]\begin{gathered} 5=\frac{1}{2}CD \\ \Rightarrow CD=5\times2=10 \end{gathered}[/tex]
Since the point B is a midpoint, it follows by definition that:
[tex]\begin{gathered} BC=AB,\text{ but AB=3} \\ \Rightarrow BC=3 \end{gathered}[/tex]
Also, E is a midpoint of AD, it follows that:
[tex]\begin{gathered} ED=\frac{1}{2}AD \\ \Rightarrow ED=\frac{1}{2}\times8=4 \end{gathered}[/tex]
The perimeter of BCDE is the sum of all sides:
[tex]P=BC+CD+ED+BE[/tex]
Substitute the length of the sides:
[tex]P=3+10+4+5=22\text{ units}[/tex]
The perimeter of BCDE is 22 units.
