17-126 $D$ is a point on the base $BC$ of an isosceles acute $\triangle ABC$, then $AD, BD$, $CD$ satisfy the relation
(A) $A D^{2}=B D^{2}+C D^{2}$.
(B) $A D^{2}>B D^{2}+C D^{2}$.
(C) $2 A D^{2}=B D^{2}+C D^{2}$.
(D) $2 A D^{2}>B D^{2}+C D^{2}$.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
[Solution] Consider the extreme case: $\triangle ABC$ is an equilateral triangle, and $D$ is the midpoint of $BC$. Then, $AD^2 > BD^2 + CD^2$.
If $D$ is at the endpoint of $BC$, we have
$$
AD^2 = BD^2 + CD^2 \text{ (note that } CD = 0 \text{), }
$$
Therefore, we can conclude that
$$
2AD^2 > BD^2 + CD^2 \text{. }
$$

Thus, the correct choice is $(D)$.