Is There Sufficient Information?

First, you must understand that $\Delta ACD\sim\Delta CDE$ . Here's one way to show this:

1) $\angle BDE\cong\angle ACB$ because they are both right angles.

2) $\angle B$ is shared by $\Delta ABC$ and $\Delta CBE$ .

3) $\Delta ABC\sim\Delta CBE$ because of AA similarity.

4) $\angle ECD\cong\angle A$ because of CASTC (Corresponding Angles of Similar Triangles are Congruent).

5) $\angle ADC\cong\angle CED$ because they are both right angles.

6) $\Delta ACD\sim\Delta CDE$ because of AA similarity.

Incidentally, all triangles in this figure are similar, and you can use the same kind of approach above to prove as such.

Now let $CD=x$ . Using corresponding parts of the similar $\Delta ACD$ and $\Delta CDE$ , we can write the proportion $\frac{100}{x}=\frac{x}{36}$ . Solving this yields $x=CD=60$ .

With the understanding that all triangles in the figure are similar, we can see that they are all proportional to a $3-4-5$ triangle. We can use this fact to solve $BC=75$ and $AB=125$ . Thus, the perimeter of $\Delta ABC$ is $\boxed{300}$ .

Let the ratio of side lengths of $\triangle ABC$ be $BC:CA:AB = a:b:c$ . Note that $a<b<c$ .

We note that $\triangle ACD$ is similar to $\triangle ABC$ , therefore, $\dfrac{CD}{CA} = \dfrac{a}{c}$ , $\quad \Rightarrow CD = \dfrac{a}{c} CA = \dfrac{100a}{c}$ .

Similarly,

$\begin{aligned} \frac{DE}{CD} & = \frac{a}{c} \\ \Rightarrow DE & = \frac{a}{c} CD = \frac{100a^2}{c^2} = 36 \\ \frac{a^2}{c^2} & = \frac{36}{100} \\ \Rightarrow \frac{a}{c} & = \frac{3}{5} \\ \Rightarrow a:b:c & = 3:4:5 \\ BC:CA:AB & = 75:100:125 \\ \Rightarrow BC+CA+AB & = 75+100+125 = \boxed{300} \end{aligned}$