A photon $Z$ of energy $\g_0$ (with respect to $mT$) gets scattered by a material particle $mT$ of rest mass $m$, i.e. $\g_0Z$ collides with $mT$ and the product is a moving particle $mX$ of rest mass $m$ and a photon $\g W$ of energy $\g$ (with respect to $mT$). Deduced from the energy-momentum conservation: $\g_0Z+mT=mX+\g W$ that $$ \g_0-\g=\frac{-\g_0\g\la Z,W\ra}m~. $$ If $\a$ is the angle of $Z$ and $W$ for $T$, then $$ \g_0-\g=\frac{2\g_0\g\sin^2(\a/2)}m~. $$ Thus the energy $\g$ of the photon after the collision is always smaller or equal to $\g_0$. wikipedia.
The conservation of energy-momentum gives us $mX=\g_0Z+mT-\g W$ and since $mX$ is a material particle of rest mass $m$ and $Z$ and $W$ are light-like and $\la Z,T\ra=\la W,T\ra=-1$ (because $\g_0=-\la\g_0Z,T\ra$ and $\g=-\la\g W,T\ra$): \begin{eqnarray*} -m^2 &=&\la mX,mX\ra =\g_0^2\la Z,Z\ra+m^2\la T,T\ra+\g^2\la W,W\ra\\ &&+2\g_0m\la Z,T\ra-2\g_0\g\la Z,W\ra-2\g m\la T,W\ra\\ &=&-m^2-2\g_0m-2\g_0\g\la Z,W\ra+2\g m~. \end{eqnarray*} It follows that $$ \g_0-\g=\frac{-\g_0\g\la Z,W\ra}m~. $$ The angle $\a$ of $Z$ and $W$ for $T$ is given by $$ 2\sin^2(\a/2)=-\frac{\la Z,W\ra}{\la Z,T\ra\la W,T\ra} =-\la Z,W\ra~. $$ Thus $T$ can determine the energy of $W$ by measuring the angle $\a$ of the two photones $Z$ and $W$.