Here is how I would discover that one needs to add, or rather subtract, $\;b\;$, without any 'magic': I would start at the most complex side, and simplify from there, as follows:
\begin{align}
& \tfrac {a+b} 2 < b \\
\equiv & \;\;\;\;\;\text{"arithmetic: multiply both sides by 2 -- to simplify the left hand side"} \\
& a+b < 2 \times b \\
\equiv & \;\;\;\;\;\text{"subtract $\;b\;$ from both sides -- to simplify both sides at the same time"} \\
& a < b \\
\end{align}
And we are already done.
Now, some call this proof 'backwards'. However, simplifying $\;\tfrac {a+b} 2 < b\;$ is a lot easier than trying to add information to $\;a < b\;$: in the above calculation, at each point there is not much choice about what to do next, while $\;a < b\;$ is a starting point which allows too much choice.
Also, note how the above does not only prove the required "if $\;a < b\;$ then $\;\tfrac {a+b} 2 < b\;$", but it proves the stronger equivalence
$$
\tfrac {a+b} 2 < b \;\equiv\; a < b
$$
for any real $\;a,b\;$.
Finally, the above "calculational" proof format (designed by Scholten and Dijkstra, and used also in the Gries-Schneider book) is concise and readable at the same time.