2005 AMC 12B 第 6 题

Drop the altitude from $C$ to line $AB$; it meets $AB$ at its midpoint $H$

$CH^2 = 7^2 - 1^2,$ and $CD^2 = CH^2 + HD^2$ with $HD = 1 + BD$

Let $H$ be the foot of the altitude from $C$ to line $AB.$ Since $\triangle ABC$ is isosceles with $AC = BC,$ $H$ is the midpoint of $AB,$ so $AH = HB = 1.$

Then $CH^2 = 7^2 - 1^2 = 48.$ Applying the Pythagorean Theorem to $\triangle CHD$ with $HD = HB + BD = 1 + BD$ gives $$8^2 = 48 + (1 + BD)^2,$$ so $(1 + BD)^2 = 16.$

Therefore $1 + BD = 4,$ which means $BD = 3.$

Thus, the correct answer is A.