It is easy to see that the triples $(1,2,6)$, $(2,3,7)$, $(3,4,8)$, $(4,5,9)$, $(5,1,10)$ cannot have the same area because if we suppose that three triangles $1,2,6$ have the same area, then quadrilateral $ABNR$ is a parallelogram and $AR \parallel BN$, a contradiction. The other cases can be proved similarly.

We consider two such cases of the set $B$:

(1) Case $B \subset \{1,2,3,4,5\}$. We can see that the triangles with indices $i, i+1, i+2$ and $i$ considered as the remainder when divided by $5$, cannot be in the same set as $B$ because if not, then one of the five above triples will appear in the set $A$, contradiction. So, we can assume that $B = \{1,2,4\}$ and then, we have $A = \{3,5,6,7,8,9,10\}$.
From the ratio of area and the corresponding base segment, we can see that $AR = RQ = QD$, $BN = NP = PD$, this leads to $PQ \parallel NR \parallel AB$ and $\frac{ER}{BR} = \frac{AR}{RQ} = 1$ or $ER = BR$ and $ER > MB$, also a contradiction.

(2) Case $B$ has $2$ triangles belonging to $\{1,2,3,4,5\}$ and $1$ triangle belonging to $\{6,7,8,9,10\}$. Without loss of generality, we suppose $6 \in B$ then $B = \{3,5,6\}$ (because of $(*)$). Then $7$ triangles have the same area: $1,2,4,7,8,9,10$.
Then $QR$ is the mid-segment of triangle $EMP$ and $QR \parallel MP$ or $AQ \parallel MP$.
Moreover, $PQ = PC$ then $AM = MC$.
Similarly, we also have $MB = ME$ or $ABCE$ is a parallelogram, which leads to
$$\frac{DQ}{DA} = \frac{PQ}{AB} = \frac{1}{3}.$$
So, we can see that $DQ = \frac{1}{3}DA$, $DQ = AR$ then $DQ = QR = RA$ and triangles $1,2,3$ have the same area, a contradiction.

Hence, the case of $7$ triangles which have the same area cannot appear and we only need to point out a case of pentagon such that $6$ triangles have the same area.

Next, we will construct the pentagon to satisfy the condition.

Construct isosceles triangle $ACE$ with $AC = AE$; on the side $AD$, $AC$, choose some points $M, N, Q, R$ in such a way that $AM = MN = NC$, $AR = RQ = QD$. Let $B, E$ be the intersections of $DN$, $CQ$ with the line $RM$. We have the pentagon $ABCDE$ satisfies the condition of the given statement with the triangles $1,2,3,4,5,6$ having the same area.

Therefore, the maximum value we need to find is $6$.