Let us look at the problem in a slightly different angle. If one know
baricentric coordinate system, it will be "obvious" why this is true.
Let $\vec{A}, \vec{B}, \vec{C}$ be any three non-collinear points,
they form the vertices of a non-degenerate triangle $\triangle ABC$.
For any point $\vec{p} \in \mathbb{R}^2$, there exists a unique
pair of real numbers $\alpha, \beta$ such that
$$\vec{p}-\vec{C} = \alpha( \vec{A}-\vec{C}) + \beta( \vec{B} - \vec{C} )
\quad\iff\quad \vec{p} = \alpha \vec{A} + \beta \vec{B} + (1 - \alpha - \beta)\vec{C}
$$
Let $\gamma = 1 - \alpha - \beta$, the triplet $(\alpha,\beta,\gamma)$ is called the baricentric coordinates for $\vec{p}$.
Furthermore, the points $\vec{p}$ lies inside or on $\triangle ABC$ if and only if $\alpha, \beta, \gamma \ge 0$
Let $h_A, h_B, h_C$ be the height of $\triangle ABC$ for corresponding vertices.
The distance between $\vec{p}$ and the sides $BC$, $CA$, $AB$ are
$h_A |\alpha|$, $h_B |\beta|$ and $h_C|\gamma|$ respectively.
The loucs for a point whose sum of distances to the 3 sides equal to $d$ is then given by:
$$h_A |\alpha| + h_B|\beta| + h_C|\gamma| = d$$
For points inside $\triangle ABC$, the problem of finding the locus is
equivalent to solving following pair of linear equations:
$$\begin{array}{rrrl}
\alpha +& \beta +& \gamma &= 1\\
h_A \alpha +& h_B \beta +& h_C \gamma &= d
\end{array}
$$
When $\triangle ABC$ is not equilateral, this pair of equations has rank 2 which
has either zero or infinite many solutions. Furthermore if $(\alpha, \beta, \gamma)$ is a solution, other solution will have the form:
$$(\alpha',\beta',\gamma') = (\alpha,\beta,\gamma) + \lambda (h_B-h_C,h_C-h_A,h_A-h_B)\quad\text{ for some } \lambda \in \mathbb{R}$$
Translate this back to points on $\mathbb{R}^2$. This mean is $\vec{p}$ is a point
inside $\triangle ABC$, the locus of point $\vec{p}'$ have same sum of distances has the form:
$$\vec{p}' = \vec{p} + \lambda\vec{u}, \quad\text{ for some }\;\lambda \in \mathbb{R}$$
i.e. the locus is a line along the direction
$\displaystyle\;\vec{u} = (h_B-h_C)\vec{A} + (h_C-h_A)\vec{B} + (h_A-h_B)\vec{C}$.
Please note that this $\vec{u}$ is independent of choice of $d$ and hence $\vec{p}$.
What this means is for all points inside $\triangle ABC$, not only the locus of same distances are all lines, all those lines are parallel to each other!
Clarifications
About the question why multiplying $\alpha$ with height $h_A$ gives us the
distance to line $BC$. For any point $p$, let $d_p$ be the distance of $p$
to the line $BC$. By definition, we have
$$\vec{p} - \vec{C} = \alpha(\vec{A}-\vec{C}) + \beta(\vec{B}-\vec{C}).$$
For any fixed $\alpha$, let $\vec{p}_0 = \vec{C} + \alpha(\vec{A}-\vec{C})$. For any point $p$ with same $\alpha$, we have
$$\vec{p} - \vec{p_0} = \beta (\vec{B} - \vec{C})$$
When viewed from $p_0$, $p$ is along the direction $\vec{B}-\vec{C}$. This means the locus of $p$ for fixed $\alpha$ is a line parallel to the side $BC$. As a result, $d_p$ is constant over such a line and $d_p$ depends only on $\alpha$.
As long as $p$ doesn't crosses the line $BC$, it is clear this dependence on $\alpha$ is linear. Notice
- When $\alpha = 0$, the line of constant $\alpha$ coincides with line $BC$, so $d_p = 0$ there.
- When $\alpha = 1$, the line of constant $\alpha$ crosses $A$, so $d_p = h_A$ there.
Combine these, we find the proportional constant is $h_A$ when $\alpha \ge 0$.
This means as long as $p$ is on the same side as $A$ with respect to line $BC$, $d_p = h_A \alpha = h_A |\alpha|$. By symmetry, $d_p = h_A |\alpha|$ on the other side too.
