$17 \cdot 89$ In $\triangle A B C$, $A B=A C=3.6, D$ is a point on $A B$  
and $A D=1.2, E$ is a point on the extension of $A C$, the area of $\triangle A E D$ equals the area of $\triangle A B C$, then $A E$ equals  
(A) 4.8.  
(B) 5.4.  
(C) 7.2.  
(D) 10.8.  
(E) 12.6.  
(10th American High School Mathematics Examination, 1959)
[Solution] Let the height from vertex $A$ to side $AC$ in $\triangle ABC$ be $h$. Also,
$$
AD = 1 \cdot 2 = \frac{1}{3} AB,
$$

Construct $DK \perp AC$ at $K$, then $DK = \frac{1}{3} h$.
Since $S_{\triangle ADE} = S_{\triangle ABC}$, we have
$$
\frac{1}{2} \cdot AE \cdot \frac{1}{3} h = \frac{1}{2} \cdot AC \cdot h,
$$

Thus, $AE = 3 \cdot AC = 10.8$.
Therefore, the answer is $(D)$.