### Home > INT1 > Chapter 8 > Lesson 8.2.3 > Problem8-120

8-120.

$\overline{BE}$ is the midsegment of $ΔACD$, shown below. What is the perimeter of $ΔACD$?

The perimeter of $ΔACD = (2)(\text{the perimeter of }ΔABE)$ because the midsegment cuts the sides in half.