ABCD  a rectangle with AB=16 units and BC = 12 units. F  is a point on AB  and  E is a point on CD  such that  AFCE is a rhombus. Then find  EF

$\begin{array}{l}{\mathrm{AC}}^2=256+144\\ \therefore {\mathrm{AC}}^2=400\\ \mathrm{AC}=20\end{array}$
Let AF = X
$AFCE$  is a rhombus
AF = FC = x
From right angled triangle BCF, ${12}^2+{\left(16-\mathrm{x}\right)}^2={\mathrm{x}}^2$
$\begin{array}{l}32\mathrm{x}=256+144=400\\ \mathrm{x}=\frac{400}{32}=\frac{25}{2}\to \left(1\right)\end{array}$
$\therefore$AFCE is a rhombus
$\therefore$ Diagonals are perpendicular
$\begin{array}{l}\mathrm{AC}=20\Rightarrow \mathrm{AO}=10\\ \text{ Let }\mathrm{EF}=2\mathrm{y}\Rightarrow \mathrm{OF}=\mathrm{y}\\ {\mathrm{x}}^2=100+{\mathrm{y}}^2\\ {\left(\frac{25}{2}\right)}^2=100+{\mathrm{y}}^2\\ {\mathrm{y}}^2=\frac{625}{4}-100\\ {\mathrm{y}}^2=\frac{625-400}{4}=\frac{225}{4}\Rightarrow \mathrm{y}=\sqrt{\frac{225}{4}}=\frac{15}{2}\\ \therefore \mathrm{EF}=2\mathrm{y}\\ =2\times \frac{15}{2}[\because \mathrm{OF}=\mathrm{OE}=\mathrm{y}]\\ \mathrm{EF}=15\mathrm{cm}.\end{array}$