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  $BD$?
Then $BM=1$. Also, $\angle CMB=\angle CMA=90^\circ$. So $CM=\sqrt{7^2-1^2}=\sqrt{48}$. So by the Pythagorean theorem, $CD=\sqrt{CM^2+DM^2}=\sqrt{48+DM^2}$. We know that $CD=8$. So we have $\sqrt{48+DM^2}=8$. Then $48+DM^2=64$. So $DM^2=16$. Then $DM=4$. Correct. $BD=3$.

