Area of a triangle

Suppose that $DX = DZ = x$ and $FX = y$ . Then Ceva's Theorem tells us that $\frac{y}{19.2} \times \frac{24}{36} \times \frac{x}{x} \; = \; 1$ and hence $y \; = \; \frac{36 \times 19.2}{24} \; = \; 28.8$ Using the Angle Bisector Theorem, we have $\frac{24}{48} \; = \; \frac{36}{2x}$ and hence $x \; = \; 36$ Thus triangle $XYZ$ has sides $48$ , $60$ and $72$ , and hence has area $|XYZ| \; = \; \sqrt{90 \times 42 \times 30 \times 18} \; = \; 540\sqrt{7}$ Hence triangle $XYE$ has area $|XYE| \; = \; \frac{24}{60}|XYZ| \; = \; 216\sqrt{7}$ and so triangle $XFE$ has area $|XFE| \; = \; \frac{y}{48}|XYE| \; = \; \frac{648}{5}\sqrt{7} \; = \; 342.8893699......$ To the nearest integer, this is $\boxed{343}$ .

By Ceva’s Theorem, we have

$(\dfrac{FY}{FX})(\dfrac{DX}{DZ})(\dfrac{EZ}{EY})=1$

Because $DX=DZ$ , we have

$\dfrac{FY}{FX}=\dfrac{EY}{EZ}$

$\dfrac{19.2}{FX}=\dfrac{24}{36}$

$FX=\dfrac{36(19.2)}{24}=28.8$

Since $\dfrac{FY}{FX}=\dfrac{EY}{EZ}$ , $FE||XZ$

Because $XE$ bisects $\angle YXZ$ , $\angle EXZ=\angle XEF$ .

Therefore triangle $XFE$ is isosceles, so $FE=FX=28.8$ .

We let $\angle YFE=\phi$ , by cosine law, we obtain

$24^2=19.2^2+28.8^2-2(19.2)(28.8)cos~\phi$

$\phi=55.7711^\circ$

It follows that

$\angle XFE=180-\phi=124.2289^\circ$

Finally the area of triangle $XFE$ is

$A_{XFE}=\dfrac{1}{2}(28.8)(28.8)(sin~124.2289)=$ $\boxed{343}$