Exercise 43 Page 369

Let's start with analyzing the given diagram.

We are told that the measure of $\overline{AC}$ is $7.$ On the diagram we can see that segments $\overline{AC}$ and $\overline{CE}$ are congruent, which means that their measures are the same. Hence, $CE$ is also $7.$ It is also given that $\overline{DC}$ measures $9.$ From the diagram we know that $\overline{DC}$ and $\overline{CB}$ are congruent segments. This allows us to conclude that $CB$ is also $9.$ Let's add this information to the diagram. Also, we can see that $\angle BCA$ and $\angle DCE$ are vertical angles. By the Vertical Angles Theorem, vertical angles are always congruent. We got that $△ABC$ and $△CDE$ have two pairs of congruent sides which form two congruent angles $\angle BCA$ and $\angle DCE.$ Therefore, by the Side-Angle-Side Theorem triangles $△ABC$ and $△CDE$ are congruent. Sides $\overline{AB}$ and $\overline{DE}$ are also congruent segments and have the same measure. Let this measure be $x.$ To find the value of $x,$ let's recall what the Triangle Inequality Theorem states. Using this theorem, we can form three inequalities for the triangle $△ABC,$ which are also true for $△CDE.$ Let's substitute the measures of the sides and solve these inequalities for $x.$

Inequality	Substitute	Solution Set
$AB+BC>AC$	$x+9>7$	$x>\text{-}2$
$AB+AC>BC$	$x+7>9$	$x>2$
$BC+AC>AB$	$9+7>x$	$x<16$

Now we can graph these solutions sets and find the common solutions.

We got that the value of $x,$ which is the measure of $\overline{AB}$ and $\overline{DE},$ must be greater than $2$ and less than $16.$ Let's use these values to find the smallest and greatest value of the perimeter of $ABCDE.$ We will find the perimeter of $△ABC$ and then multiply by $2,$ because $ABCDE$ consists of two triangles of the same perimeter. Therefore, the range for the possible perimeters of $ABCDE$ is from $36$ to $64.$