The claim is that if triangles $ABC$ and $XYZ$ are perspective from a point $P$ then they are perspective from a line (meaning that the three points $$AB\cdot XY\qquad BC\cdot YZ\qquad CA\cdot ZX$$ are collinear) and conversely.

Since no three of $A,B,C,P$ are collinear, we can lay down homogeneous coordinates such that $$A=(1,0,0)\qquad B=(0,1,0)\qquad C=(0,0,1)\qquad P=(1,1,1)$$ By hypothesis, there are scalars $p,q,r$ such that $$X=(1,p,p)\qquad Y=(q,1,q)\qquad Z=(r,r,1)$$ The equation for a line through $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is $$(y_1z_2-z_1y_2)x+(z_1x_2-x_1z_2)y+(x_1y_2-y_1x_2)z=0\;,$$ giving us equations for six lines: \begin{eqnarray*} AB&:&z=0 \\ BC&:&x=0 \\ CA&:&y=0 \\ XY&:&(pq-p)x+(pq-q)y+(1-pq)z=0 \\ YZ&:&(1-qr)x+(qr-q)y+(qr-r)z=0 \\ ZX&:&(rp-p)x+(1-rp)y+(rp-r)z=0 \end{eqnarray*}whence \begin{eqnarray*} AB\cdot XY&=&(pq-q,-pq+p,0)\\ BC\cdot YZ&=&(0,qr-r,-qr+q)\\ CA\cdot ZX&=&(-rp+r,0,rp-p). \end{eqnarray*}As claimed, these three points are collinear, since the determinant $$ \left| \begin{array}{ccc} pq-q & -pq+p & 0 \\ 0 & qr-r & -qr+q \\ -rp+r & 0 & rp-p \end{array}\right| $$ is zero. (More precisely, all three points are on the line $$p(q-1)(r-1)x+(p-1)q(r-1)y+(p-1)(q-1)rz=0\;.)$$

Since the hypotheses are self-dual, the converse is true also, by the principle of duality.