Number of tangential quadrilaterals with distinct sidelengths chosen in set $\{1,2,\ldots, 8\}$ 
There are $8$ line segments of length $1,2,3,\ldots,8$ $\textrm{units}$. How many quadrilaterals can be made from these line segments such that circles can be inscribed in the quadrilaterals made$?$


I applied Pitot theorem. I first fixed on side as $8$ then found all possible combinations. Then, I fixed $7$ and counted all possible combinations excluding those which include $8$ and so on. I am listing all the combinations that I found.
$(8,1,7,2),(8,1,6,3),(8,1,5,4),(8,2,7,3),(8,2,6,4),(8,3,7,4),(8,3,6,5),(8,4,7,5)$
$(7,1,6,2),(7,1,5,3)(7,2,6,3),(7,2,5,4),(7,3,6,4),(7,4,6,5),(6,1,5,2),(6,1,4,2)$
$(6,2,5,3),(6,3,5,4),(5,1,4,2),(5,2,4,3),(4,1,2,3)$
I found $21$ combinations in all. Is my method and my answer correct$?$ Is there any other elegant way to arrive to the correct answer$?$
Any help is greatly appreciated. There is a similar question this on this site, stating the same problem but the solution given there doesn't give any answer or any elegant method.

A good way to proceed is to take the twenty-eight possible pairs of side lengths and classify them according to the sum of the lengths:
$\{1,2\}$
$\{1,3\}$
$\{1,4\},\{2,3\}$
$\{1,5\},\{2,4\}$
$\{1,6\},\{2,5\},\{3,4\}$
$\{1,7\},\{2,6\},\{3,5\}$
$\{1,8\},\{2,7\},\{3,6\},\{4,5\}$
$\{2,8\},\{3,7\},\{4,6\}$
$\{3,8\},\{4,7\},\{5,6\}$
$\{4,8\},\{5,7\}$
$\{5,8\},\{6,7\}$
$\{6,8\}$
$\{7,8\}$
Now select any pair which will represent two opposite sides and join it with another pair having the same sum (which will be the other pair of opposing sides). Clearly there is no match for sums of $3,4,14,15$. For the rest of the cases, if you can choose from $m$ pairs with a given sum you have $\binom{m}{2}=m(m-1)/2$ choices for the two opposing side-pairs that make up the quadrilateral. Thus one choice with sums of $5$, one with sums of $6$ (both $m=2$), three apiece with sums of $7$ and $8$ (both $m=3$), etc., and the sum of these all these combination counts will be $22$ as indicated in the comments.
