Hello, MATNTRNG!
This one is surprisingly easy . . .
The Indian mathematician Brahmagupta (7th century) found that
the area of a cyclic quadrilateral with sides $\displaystyle a,b,c,d$ is given by:
. . $\displaystyle A \;=\;\sqrt{(p-a)(p-b)(p-c)(p-d)}\;\text{ where }p = \frac{1}{2}(a+b+c+d)$
Show that Heron's Formula for the area of a triangle with sides $\displaystyle a,b,c\!:$
. . . $\displaystyle A \;=\;\sqrt{p(p-a)(p-b)(p-c)}\:\text{ where }p \:=\:\frac{1}{2}(a+b+c)$
follows from Brahmagupta's result.
$\displaystyle \text{Let }\,d = 0.$