Claim 1: A parallelogram $ABCD$ can be tiled (covered without overlap except for edges) with equilateral triangles only if its angles are $60$ and $120$ degrees.
Proof: WLOG assume that $\angle A=\angle C\leq \angle B=\angle D$. Note that $\angle A+\angle B=180^\circ$. Now the tip $A$ of the parallelogram must be covered by some equilateral triangle, meaning that $A$ must be the vertex of at least $1$ equilateral triangle. Thus $\angle A\geq 60^\circ$. If there are two distinct equilateral triangles with vertex $A$, then $\angle A\geq 120^\circ$, contradicting the fact that $\angle A \leq \angle B$ and $\angle A+\angle B=180^\circ$. Thus $\angle A=60^\circ$ and $\angle B=120^\circ$ necessarily.
Claim 2: Let $ABCD$ be a parallelogram with $\angle A=\angle C=60^\circ$ and $\angle B=\angle D=120^\circ$ such that $[AD]=[BC]=1$ and $[AB]=[CD]=n$ for some natural number $n\geq 1$. Then the 'minimal tiling by equilateral triangles' for $ABCD$ is achieved with $2n$ triangles of side-length $1$.
Proof: Induction. When $n=1$, it is easy to see that we need at least $2$ triangles. When going from $n$ to $n+1$, note that the largest possible triangles we could use in the tiling cannot have side-length strictly greater than $1$. The area of $ABCD$ is easily seen to be $(n+1)\tfrac{\sqrt{3}}{2}$ and the area of an equilateral triangle of side-length at most $1$ is at most $\tfrac{\sqrt{3}}{4}$. The conclusion is now immediate.
Claim 3: Let $ABCD$ be a parallelogram with $\angle A=\angle C=60^\circ$ and $\angle B=\angle D=120^\circ$. Let, as in the picture from the post, $E\in[AB]$ and $F\in[CD]$ be such that $DE$ is the bisector of $\angle ADC$ and $BF$ is the bisector of $ABC$. Assume that we know that the minimum number of equilateral triangles needed to cover the parallelogram $EBFD$ is already known and it is $N$ for some natural $N$. Then the minimal tiling of $ABCD$ will have $N+2$ triangles.
Proof: It is trivial to see that $\triangle ADE$ along with $\triangle CBF$ and the minimal covering for $EBFD$ achieves the $N+2$ covering for $ABCD$. But can we do better? The only attempt is to have $D'\in [AD]$ and $E'\in[AE]$ such that $\triangle AD'E'$ is equilateral and similar for $\triangle CB'F'$. This means that we are looking for a covering with less than $N$ triangles for the hexagon $D'E'BB'F'D$. Let us consider the trapezoid $D'E'ED$ which lies inside this hexagon. Such a 'smaller' cover of the hexagon cannot admit a subcover that covers $D'E'ED$ without excess since then we could split the hexagon cover into a cover for $D'E'ED$, one for $BB'F'F$ and one for $EBFD$ and this will definitely have more than $N$ triangles.
not rigurous!!: Since the angles of the trapezoid are multiples of $60^\circ$, the sides of the triangles that form this hypothetical subcover of $D'E'ED$ with excess must be parallel to the sides of the trapezoid. If we believe this, the only hope to get a smaller cover for the hexagon is if the excess from covering $D'E'ED$ that 'spills' into $EBFD$ will be a union of equilateral triangles. But then, we will definitely need at least $N$ triangles to cover the hexagon, so we are not finding a cover with less than $N+2$ triangles for $ABCD$.
Claim 4: Let $ABCD$ be a parallelogram with $\angle A=\angle C=60^\circ$ and $\angle B=\angle D=120^\circ$ such that $[AD]=[BC]=m$ and $[AB]=[CD]=n$ for some natural numbers $m<n$ that are coprime. Then the 'minimal tiling by equilateral triangles' for $ABCD$ is achieved with a number of equilateral triangles equal to $2$ times the number of steps in the 'Euclidean algorithm by subtraction' applied to the pair $(n,m)$.
Proof: To clarify what is meant, consider for example the 'Euclidean algorithm by subtraction' for the pair $(17,12)$. This will look like $(17,12)\rightarrow (12,5) \rightarrow (7,5)\rightarrow (5,2)\rightarrow(3,2)\rightarrow (2,1) \rightarrow (1,1)$. It is easy to see that this will correspond to the proposed solution by contracting multiple steps where the role of numerator and denominator does not flip, e.g., the two steps $(12,5)$ and $(7,5)$ can be condensed into $\left\lfloor \tfrac{12}{5}\right\rfloor$.
Again we will proceed by a sort of induction or descent along the steps of the 'Euclidean algorithm by division'. Crucially, since $m$ and $n$ are coprime, this will always terminate in $(1,1)$ and, in fact, the last steps $(q,1)\rightarrow\dots\rightarrow(1,1)$ are already covered by Claim 2.
The step case is covered (although not very formally) by Claim 3.
