19. As shown in the figure, in $\triangle \mathrm{ABC}$, if $A C=\frac{\sqrt{5-A B}+\sqrt{2 A B-10}+A B+1}{2}$, then the range of the median $\mathrm{AD}$ on side $\mathrm{BC}$ is ( )
(A) $2<A D<8$
(B) $0<A D<8$
(C) $1<A D<4$
(D) $3<A D<5$
Answer: (C)

Solution: According to the problem, we have
$$
\left\{\begin{array}{l}
5-A B \geq 0 \\
2 A B-10 \geq 0
\end{array},\right.
$$

Solving the two inequalities, we get
$$
A B \geq 5 \text { and } A B \leq 5 \text {, }
$$

Therefore,
$$
A B=5 \text {, }
$$

Thus,
$$
A C=\frac{\sqrt{5-A B}+\sqrt{2 A B-10}+A B+1}{2}=3 .
$$

As shown in the figure, extend $\mathrm{AD}$ to $\mathrm{E}$, such that $A D=D E$, and connect $\mathrm{BE}$.
Since $D$ is the midpoint of $\mathrm{BC}$,
we have
$$
B D=C D \text {, }
$$

Also,
$$
\angle A D C=\angle B D E, A D=D E,
$$

Therefore,
$$
A D C \cong E D B \text { (SAS) }
$$

Thus,
$$
A C=B E \text { . }
$$

In $\triangle \mathrm{ABE}$, we have
$$
A B-B E<A E<A B+B E \text {, }
$$

So,
$$
2<A E<8 \text {, }
$$

That is,
$$
2<2 A D<8 \text {, }
$$

Therefore,
$$
1<A D<4 \text {. }
$$

Hence, the answer is (C).