Is there a derivation of the Early effect?
If \$A\$ is the angle at A, and \$I_{CA}\$ is the current at \$V_{CE}=0\$ 
which approximately equals \$I_{C}\$ (with no Early effect), then

\$\tan(A) = I_{CA}/V_A\$

If \$\Delta I_{CA}\$ is the increase in the value of \$I_c\$ due to the Early effect then also,

\$\tan(A) = (I_{CA} + \Delta I_C ) / (V_A+V_{CE})\$

Now solve for \$(I_{CA}+ \Delta I_C)\$ you will get the above equation.