AMC10 2005 10B, Problem 10

In $\triangle ABC$, we have $AC=BC=7$ and $AB=2$. Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD=8$. What is $BD$?

Solution: Draw height $CH$ (Perpendicular line from point C to line AD). We have that $BH=1$. By the Pythagorean Theorem, $CH=\sqrt{48}$. Since $CD=8$, $HD=\sqrt{8^2-48}=\sqrt{16}=4$, and $BD=HD-1$, so $BD=\fbox{3}$.

Answer: 3