Another problem from my chinese math practice book

$\frac{1}{2}\times AB \times BF=24$

$\frac{1}{2}\times 8\times BF=24$

$4BF=24$

$BF=6$

Considering $\angle ABF=90$ degrees, $AB=8$ and $BF=6$ tells us that $AF=10$ employing Pythagoras' Theorem and $AD=AF=10$ . Moreover, $BC=AD=10$ by virtue of the fact that $ABCD$ is a parallelogram. Then $CF=BC-BF=10-6=4$ .

If $CE=x$ , then $DE=CD-CE=8-x$ . Considering that $AD=AF$ and $AE$ bisects $DF$ , we eventually see that $ADE$ and $AEF$ are congruent. Then $EF=DE=8-x$ . Now,

$CF^{2}+CE^{2}=EF^{2}$

$4^2+x^2=(8-x)^2$

$16+x^2=64-16x+x^2$

$16x=48$

$x=\frac{48}{16}=\boxed{3}$

@Anthony Hong . I have a favour. I really love this Chinese maths book. Can you send this to me at least through pdf to my email hamzjeorge@gmail.com Please I really love that book

Did you take the picture from the front camera?