Solution:

For the purpose of this proof, we will consider a vertex to be any point which is a corner of at least one of the triangular pieces. Define an edge to be any line segment between two vertices, which is part of a side of a triangular piece but does not pass through any other vertex. Note that each vertex must be one of the following types:
- Type $A$: incident with only 2 edges at $60^{\circ}$.
- Type $B$: incident with exactly 4 edges at angles $60^{\circ}, 60^{\circ}, 60^{\circ}, 180^{\circ}$.
- Type $C$: incident with exactly 6 edges forming six $60^{\circ}$ angles.

There can be no other types of vertex, because all the angles must be $60^{\circ}$ or $180^{\circ}$ (or $300^{\circ}$ in the corners of the original large triangle). We will now colour each edge-end either green or blue, as shown in the diagram (the green edge-ends are also a bit thicker).

An edge-end is coloured green if it touches a Type $B$ vertex along the $180^{\circ}$ angle, otherwise it is coloured blue. Now observe that there must be exactly 3 vertices of type $A$ (the three corners of the original large triangle before it was partitioned). If there are $x$ vertices of type $B$ and $y$ vertices of type $C$ then this makes a total of
$$(6 + 2x + 6y)$$
blue ends, but only $$(2x)$$ green ends.

Note also that the 6 edges with an endpoint at a type $A$ vertex must all have their other end being green. There are more blue edge-ends than green edge-ends, so it can't be the case that every blue edge-end is connected to a green edge-end. There must therefore be an edge such that both of its ends are blue. Since neither endpoint of such an edge is a type $A$ vertex, we can conclude that all four of the angles at the endpoints of this double blue-ended edge must be $60^{\circ}$.

Therefore this edge is a shared side of two triangular pieces. These two triangular pieces must therefore be the same size. $\square$