2011 AMC 10B Problem 9

Let's first see what is in the figure. Triangle $ABC$ has $AC = 3$, $BC = 4$, $AB = 5$, with the right angle at $C$. Point $D$ is on $AB$, point $E$ is on $BC$, and $DE$ makes a right angle with $AB$.

1. The two triangles have the same shape. Triangle $EBD$ has a right angle at $D$. Triangle $ABC$ has a right angle at $C$. Both triangles also use the angle at $B$. Two triangles with two equal angles have the same shape (they are similar).
2. Match up the corners. Right angle goes with right angle, so $D$ goes with $C$. $B$ goes with $B$. That leaves $E$ with $A$.
3. So side $BD$ (from $B$ to the right angle) matches side $BC$ (from $B$ to the right angle), which is $4$.
4. Turn the area ratio into a side ratio. If a triangle's sides are multiplied by $k$, its area is multiplied by $k \times k$. Here the area is multiplied by $\frac{1}{3}$, so $k \times k = \frac{1}{3}$ and $k = \frac{1}{\sqrt{3}}$.
5. So
$$BD = 4 \times \frac{1}{\sqrt{3}} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{4\sqrt{3}}{3}}$.

When you draw a line from a side of a right triangle that makes a right angle, you often cut off a smaller copy of the big triangle. For shapes that are copies of each other, area grows like the square of the side, so a side ratio of $k$ means an area ratio of $k^2$. Be careful which sides match: $BD$ lies along the long side $AB$ of the big triangle, but in the small triangle it plays the role of $BC$.

Call $BD = x$. In triangle $EBD$, the side $DE$ compares to $BD$ the same way $AC$ compares to $BC$, so $DE = \frac{3}{4}x$. The area of $EBD$ is $\frac{1}{2} \times x \times \frac{3}{4}x = \frac{3}{8}x^2$. The big triangle has area $\frac{1}{2} \times 3 \times 4 = 6$, and a third of that is $2$. So $\frac{3}{8}x^2 = 2$, which gives $x^2 = \frac{16}{3}$ and $x = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}$.