Michaelis menten constant motivation behind 1/2?
To see why the $K_m$ is equivalent to the substrate concentration at which the rate is equal to $V_{max}/2$, we need to look at the derivation of the Michaelis-Menten, which is usually included in detail in any introductory biochemistry book. Here is a brief version.
We will consider an enzyme-catalyzed reaction described by the following scheme:
$$\ce{E + S &lt;=&gt;[k_1][k_{-1}] ES -&gt;[k_2] E + P},$$
where &quot;E&quot; is enzyme, &quot;S&quot; is substrate and &quot;P&quot; is product.
To get an expression for the rate, we start by noting that the rate of formation of product is $k_2[\ce{ES}]$, so we need to find an expression for $[\ce{ES}]$ using known quantities. Next, we make a steady-state approximation, setting $\frac{d[\ce{ES}]}{dt}=0$, so we have
$\displaystyle k_1[\ce{E}][\ce{S}]-(k_{-1}+k_2)[\ce{ES}]=0$
$\displaystyle \implies [\ce{ES}]=\frac{k_1}{k_{-1}+k_2}[\ce{E}][\ce{S}].$
To finish, we note that when the reaction has not consumed very much substrate, we can approximate $[\ce{S}]$ as equal to the starting substrate concentration $[\ce{S}]_0$. We also know that the total amount of enzyme in the reaction must be equal to the sum of the concentrations of E and ES, ie $[\ce{E}]+[\ce{ES}]=[\ce{E}]_{Total}$. We already have an equation for $[\ce{ES}]$ in terms of $[\ce{E}]$, so with some algebra, we can find that
$\displaystyle [\ce{ES}]=\frac{k_1[\ce{E}]_{Tot}[\ce{S}]}{k_{-1}+k_2+k_1[\ce{S}]}$.
Combining everything, we get
rate of formation of $\displaystyle \mathrm{P} =k_2[\ce{ES}]=\frac{k_2k_1[\ce{E}]_{Tot}[\ce{S}]_0}{k_{-1}+k_2+k_1[\ce{S}]_0}$.
In order to isolate $[\ce{S}]_0$ in the denominator, we divide the numerator and denominator by $k_1$:
rate $\displaystyle =\frac{k_2[\ce{E}]_{Tot}[\ce{S}]_0}{\frac{k_{-1}+k_2}{k_1}+[\ce{S}]_0}$,
and our final simplification is to define the unwieldy $\frac{k_{-1}+k_2}{k_1}$ as $K_m$ and $k_2[\ce{E}]_{Tot}$ as $V_{max}$:
rate $\displaystyle =\frac{V_{max}[\ce{S}]_0}{K_m+[\ce{S}]_0}$.
Getting back to your original question: Note that we never explicitly defined $K_m$ to be the substrate concentration at half-maximal rate. It simply resulted from our choice to isolate the substrate term in the denominator. However, that was not a completely arbitrary choice.
If we look at $K_m=\frac{k_{-1}+k_2}{k_1}$, we can see that if $k_{-1}$ is large and $k_2$ is small, $K_m$ approaches $\frac{k_{-1}}{k_1}$, which is the equilibrium constant for dissociation of the substrate from ES. For that reason, $K_m$ is often treated as an apparent measure of the affinity of the enzyme for the substrate, and that is an additional reason why we choose to define $K_m$ this way. Always remember that it is not actually the dissociation constant, and in some reactions will be quite far off from the actual $K_d$.