$22 \cdot 8 AB, BC, CD$ are three line segments not in the same plane. The midpoints of $AB, BC$, and $CD$ are $P, Q, R$, respectively, and $PQ=2, QR=\sqrt{5}, PR=3$. Then the angle formed by the lines $AC$ and $BD$ is
(A) $60^{\circ}$.
(B) $30^{\circ}$.
(C) $90^{\circ}$.
(D) $120^{\circ}$.
(China Beijing High School Mathematics Competition, 1986)
[Solution] As shown in the figure, $B D / / Q R, A C / / P Q$.
Therefore, the degree of the angle formed by $A C, B D$ is equal to the degree of $\angle P Q R$. However, in $\triangle P Q R$,
$$
P Q^{2}+Q R^{2}=2^{2}+(\sqrt{5})^{2}=3^{2}=P R^{2},
$$

Thus,
$$
\angle P Q R=90^{\circ} \text {. }
$$

Therefore, the angle formed by $A C, B D$ is $90^{\circ}$.
Hence, the answer is $(C)$.