Hard geometry problem: given a triangle with sides 16, 30, 34, find the area of the triangle introduced by the inscribed circle You are given a triangle $\triangle PQR$ with sides $16, 30, 34$. Let the incircle touch the sides of $\triangle PQR$ at $X,Y,$ and $Z$. Given that the ratio $[XYZ]/[PQR]$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers, find $m+n$.
What I did first was to draw the diagram and I noticed that $PX=PZ, QX=QY, RY=RZ$ due to two intercepting lines both being tangent to a circle. Since triangles $PXZ, QXY, RYZ$ are isosceles I bisected them and got another right triangle after I tried to do similar triangles but got a little stuck.
My other friend used "barycentric coordinates", which I thought was completely unnecessary.
Any help would be appreciated.

John Omeilan has provided an excellent solution. However the answer can be simpler if we observe that $\Delta PQR$ is a right angled triangle.

Since $16^2+30^2=34^2$, $\Delta PQR$ is a right angled triangle and hence  $ [PQR] = \frac{16 \times 30}{2}=240$
$s=\frac{p+q+r}{2}=\frac{30+16+34}{2}=40$
Let the radius of the inscribed circle be $x$
From $xs=[PQR] =240$, we have
$$x=\frac{240}{40}=6$$
Notice that $\sin R=\frac{16}{34}=\frac{8}{17}, 
\sin P=\frac{15}{17}$
and $\sin Q=\frac{34}{34}=1$
Also note that $\angle YOZ=180^o-R$.
Hence $ [YOZ]=\frac{1}{2}x^2\sin(180^o-R)=\frac{1}{2}\times 6^2 \times \sin R=18 \times \frac{8}{17}$
Similarly $[ZOX]=\frac{1}{2}x^2 \times \frac{15}{17}=18 \times \frac{15}{17}$
and $[XOY]=\frac{1}{2}x^2=\frac{1}{2}\times 6^2=18$
Hence $[XYZ]=18 \times \left(\frac{8}{17}+\frac{15}{17}+1 \right)=\frac{720}{17}$
Thus $\frac{[XYZ]}{[PQR]}=\frac{\frac{720}{17}}{240}=\frac{3}{17}$
