7.2. Consider triangle $ABC$ with obtuse angle $ABC$. Points $D$ and $E$ lie on side $AC$ such that $AD \equiv DE \equiv EC$, and $BE \perp AB$ and $BD \perp BC$. Determine the length of side $BC$ given that the perimeter of triangle $ABC$ is $(12+8 \sqrt{3}) \mathrm{cm}$, and $\frac{AB}{AC}=\frac{\sqrt{3}}{3}$.
## Solution:

From $\frac{A B}{A C}=\frac{\sqrt{3}}{3}$, we consider $A B=\sqrt{3} x$ and $A C=3 x$. Since $A D \equiv D E \equiv E C$ and $A C=3 x$, it follows that $A D=D E=E C=x$.

Consider the right triangle $A B E$, with $B E \perp A B$. Since $A D=D E$, it follows that $B D$ is the median corresponding to the hypotenuse. Thus, $B D=\frac{A E}{2}=x$.

Consider the right triangle $D B C$, with $B D \perp B C$. Since $D E=E C$, it follows that $B E$ is the median corresponding to the hypotenuse. Thus, $B E=\frac{C D}{2}=x$.

Consider the right triangles $A B E$ and $D B C$. Since $A E=2 x=C D$ and $B D=x=B E$, it follows that $\triangle A B E \equiv \triangle C B D$ by the hypotenuse-leg criterion. Thus, $A B \equiv B C$. Therefore, $B C=\sqrt{3} x$.

Since $\quad P_{\triangle A B C}=12+8 \sqrt{3}, \quad$ we get $\quad \sqrt{3} x+\sqrt{3} x+3 x=12+8 \sqrt{3}, \quad x(3+2 \sqrt{3})=12+8 \sqrt{3}$, $x=\frac{12+8 \sqrt{3}}{3+2 \sqrt{3}}=4$. Since $B C=\sqrt{3} x$ and $x=4$, we get $B C=4 \sqrt{3} \mathrm{~cm}$.

Answer: $B C=4 \sqrt{3} \mathrm{~cm}$.