Thinking about the Twin Paradox, what would a Twin Traveling at .8c, calculate as the amount his EBT aged at the mid-point in a 4 lightyear trip?
I find it easier to just write down the coordinates of the events of interest.
Given coordinates $(t, x)$ in the frame of the Earth and choosing units such that $c=1$, the Lorentz transform to the frame of the traveller is $(t', x')$, where
$$\begin{align}
t' &amp;= \gamma(t - \beta x) \\
x' &amp;= \gamma(x - \beta t)
\end{align}$$
In your scenario, $\beta = \frac{v}{c} = \frac{4}{5}$, giving $\gamma = (1 - \beta^2)^{-1/2} = \frac{5}{3}$. The relevant events, in the reference frame of the Earth, are then:

Traveller leaves Earth at $(0, 0)$.
Traveller reaches halfway point at $(\frac{5}{2}, 2)$.
Earth is at the origin for all $t$, giving coordinates $(t, 0)$.

Transforming these into the traveller's frame, these become:

Traveller leaves Earth at $(0, 0)$.
Traveller reaches halfway point at $(\frac{3}{2}, 0)$ — as expected, the space coordinate is zero because the frame is moving with the traveller.
Earth is located at $(\frac{5}{3}t, -\frac{4}{3}t)$ — note this is expressed in terms of $t$, the time coordinate in the reference frame of the Earth. As expected, the Earth falls behind the traveller as time advances.

Equating the time coordinates in the traveller's frame gives $\frac{5}{3}t = \frac{3}{2}$, which gives $t = \frac{9}{10}$ years for the time that has passed on Earth, as viewed in the traveller's frame (who would have experienced a duration of $\frac{3}{2}$ years up to this point).