Let A (1, 1), B(4, 2), C(9, 3) be the vertices of a triangle ABC. A parallelogram AFDE is drawn with vertices D, E, F on the line segment BC, CA, AB respectively. The maximum area of the parallelogram is 

Let D divides BC in the ratio $\lambda :1.$ Then
$D=\left(\frac{4+9\lambda }{1+\lambda },\frac{2+3\lambda }{1+\lambda }\right),E=\left(\frac{1+9\lambda }{1+\lambda },\frac{1+3\lambda }{1+\lambda }\right)$
Area of $AFDE=2\mathrm{△}AED$
$(1+\lambda {)}^2\mathrm{\Delta }=\left|\begin{array}{ccc}1& 1& 1\\ 1+9\lambda & 1+3\lambda & 1+\lambda \\ 4+9\lambda & 2+3\lambda & 1+\lambda \end{array}\right|$
Applying $R_3\to R_3-(\lambda +1)R_1$
$\text{ and }{R}_2\to R_2-(\lambda +1)R_1\text{, we get }$
$(1+\lambda {)}^2\mathrm{\Delta }=\left|\begin{array}{ccc}1& 1& 1\\ 8\lambda & 2\lambda & 0\\ 3+8\lambda & 1+2\lambda & 0\end{array}\right|$

Applying $R_3\to R_3-R_2\text{, we get }$
$\begin{array}{r}(1+\lambda {)}^2\mathrm{\Delta }=\left|\begin{array}{ccc}1& 1& 1\\ 8\lambda & 2\lambda & 0\\ 3& 1& 0\end{array}\right|=2\lambda \\ \therefore \mathrm{\Delta }=\frac{2\lambda }{(\lambda +1{)}^2}=\frac{2}{{\left(\sqrt{\lambda }+\frac{1}{\sqrt{\lambda }}\right)}^2}\le \frac{2}{4}=\frac{1}{2}\end{array}$