$16 \cdot 19$ On a plane, given three non-collinear points $A, B, C$, construct a line $l$ such that the distances from points $A, B, C$ to line $l$ are in the ratio $1: 1: 2$ or $1: 2: 1$ or $2: 1: 1$. The number of such lines is
(A) 3.
(B) 6.
(C) 9.
(D) 12.
(E) 15.
(China Junior High School Mathematics Correspondence Competition, 1985)
[Solution] Connect $A B, B C, C A$, extend $A B$, take $B E=$ $A B$; extend $A C$, so that $C F$ 
$=A C$. Take the midpoint $D$ of $B C$. Connect $E D, F D$, and extend them respectively to intersect $A C$ at $H$, and $A B$ at $G$.

Then the lines $E H, F G, G H, E F$ are all at a distance ratio of $1: 1: 2$ from points $B$, $C$, $A$ respectively.

Similarly, taking the midpoints of $A B, A C$, another 8 lines that meet the criteria can be obtained, thus there are 12 lines in total that meet the requirements.
Therefore, the answer is $(D)$.