Consider a triangle $ABC$ and a point $P=(x:y:z)$(in Barycentric coordinates). If $P^t$ is the isotomic conjugate of $P$ prove that
$$P^t=\left(\frac{1}{x}:\frac{1}{y}:\frac{1}{z}\right)$$ Note that $$( u : v : w)=\left(\frac{u}{u+v+w} , \frac{v}{u+v+w}, \frac{w}{u+v+w}\right)$$
Now since
$$x=\frac{[PBC]}{[ABC]}, \quad y=\frac{[PCA]}{[BCA]}, \quad z=\frac{[PAB]}{[CAB]}$$
We should have
$$\frac{[ABC]}{[P^tBC]}=\frac{[PBC]}{[ABC]}\text{ etc...}$$
But I have no idea how to prove it.
Here is the diagram 
I will write as usual $(x,y,z)$ for the (normed) barycentric coordinates of a point if $x+y+z=1$, else $(x:y:z)$ is defined if the sum $S:=x+y+z\ne 0$, and this point corresponds to $(x/S,y/S,z/S)$.
Start with $P=(x,y,z)$. Then $$ \begin{aligned} G &= (0:y:z)\ , & G'&=(0:z:y)=\left(0:\frac 1y:\frac 1z\right)\ ,\\ H &= (x:0:z)\ , & H'&=(z:0:x)=\left(\frac 1x:0:\frac 1z\right)\ ,\\ I &= (x:y:0)\ , & I'&=(y:x:0)=\left(\frac 1x:\frac 1y:0\right)\ ,\\ \end{aligned} $$ and it is clear that the claimed point $$ \left(\frac 1x:\frac 1y:\frac 1z\right) $$ is on $AG'$, on $BH'$, and on $CI'$.
$\square$