$17 \cdot 72$ On the hypotenuse $AB$ of the right triangle $ABC$, another right triangle $ABD$ is constructed with $AB$ as the hypotenuse. If $BC=1, AC=b$, and $AD=2$, then $BD$ equals
(A) $\sqrt{b^{2}+1}$.
(B) $\sqrt{b^{2}-3}$.
(C) $\sqrt{b^{2}+1}+2$.
(D) $b^{2}+5$.
(E) $\sqrt{b^{2}+3}$.
(3rd American High School Mathematics Examination, 1952)
[Solution] Let $B D=x$, by the Pythagorean theorem we have
$$
x^{2}+2^{2}=A B^{2}=1+b^{2} \text {, }
$$

thus
$$
x=\sqrt{b^{2}-3} \text {. }
$$

Therefore, the answer is $(B)$.