Prove that $DG+HE=GH$ for the given figure. 
What I could gather:
$$\measuredangle ODB=\measuredangle OFC=\measuredangle OEC =90 \text{ degrees}$$
$$OD=OF=OE=r$$
$$\measuredangle ODE=\measuredangle OED$$
$$\measuredangle DOE=2\measuredangle DME$$

 A: This is inversion proof (not projective!) with respect to circle which center is $M$ and $r= MD = ME$.
Let $FM$ cuts $DE$ at $T$. It is enough to show that $H$ halves $TE$.
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Since each $D$ and $E$ maps to it self, the circle $MDE$ maps to line $DE$ and so inversion swaps $K$ and $H$, where $K$ is intersection of line $MC$ and circle $MDE$. Also, this inversion swaps $F$ and $T$.
Since $\triangle CEK \sim \triangle CME \Longrightarrow \displaystyle{KE\over ME} = {CK\over CE} \;\;(1)$.
Similary  $\triangle CFK \sim \triangle CMF \Longrightarrow \displaystyle{KF\over MF} = {CK\over CF} \;\;(2)$.
Now remeber that if $X',Y'$ are images of $X,Y$ we have $$X'Y' = XY \cdot {r^2\over XM\cdot YM}$$
So $$ HE = KE \cdot {r^2\over KM\cdot EM} \stackrel{(1)}{=}{CK\over CE} \cdot {r^2\over KM}$$
and
So $$ HT = KF \cdot {r^2\over KM\cdot FM} \stackrel{(2)}{=}{CK\over CF} \cdot {r^2\over KM}$$
Since $CE = CF$ we have $HE= HT$. q.e.d.
A: Projective solution: 
Let $t$ be a tangent in $M$. Then $t||DE$. Let $CM$ cuts segment $FE$ in $O$. Now we have:
\begin{eqnarray*}
  D(E,T;H, \infty) &=&  D(ME,MT;MH, M\infty)\\
   &=&  D(ME,MF;MK, MM)\\
   &=& D(E,F;K,M) \\
   &=& D(EE,EF;EK,EM) \\
   &=& D(EC,EO;EK,EM) \\
    &=& D(C,O;K,M) \\
&=&-1
\end{eqnarray*}
The last one is true since $O$ is on polar of $C$, so they are
harmonical conjugate.
So $H$ halves $ET$ since $D(E,T;H, \infty)=-1$. In the same manner we prove that $G$ halves $DT$ and we are done.
