Given diagonals in a parallelogram. Find sides. Given a parallelogram with $d_1 = AC = 26$ cm, $d_2 = BD = 18$ cm and $\sin \displaystyle \angle AOD = \frac{12}{13}$.
Find $AB = a$ and $AD = b$.

What I did in order to solve it

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*Using the formula for the area

$S = \frac{d_1d_2\sin \displaystyle \Phi}{2} = \frac{26 \times 18 \times 12/13}{2} = 216$

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*Using Heron's formula for $\triangle BOC$ to find $b$
Defining $OC = a = 26 / 2 = 13$, $OB = b = 18 / 2 = 9$, $BC = c$
$p = \frac{a + b +c}{2} = \frac{13 + 9 + c}{2} = \frac{22 + c}{2}$
$S = \sqrt{p(p-a)(p-b)(p-c)}$
I believe when I find $b$, I will able to find $a$ too, but kinda stuck at that point.

You already noted that the diagonals are divided by half in the center. However, you're making it far too complicated by using the area.
Instead, take a look at the triangle $\triangle AOD$. You already know the length of two sides, $a=9$ and $c=13$ and their inclosed angle $\beta=\operatorname{arcsin} \frac{12}{13}\approx 67.38°$. With the help of the cosine theorem you get:
$$
b^2=a^2+c^2-2ac\cdot\cos \beta \\ 
b = \sqrt{81+169-2\cdot 9 \cdot 13\cdot\cos\left(\operatorname{arcsin} \frac{12}{13}\right)}=4\sqrt{10}\approx 12.65
$$
By using $\angle AOB = 90°-\angle AOD$ you can calculate the length of $a$ analogously within the triangle $\triangle AOB$.
