Semiperimeter

The semiperimeter on a figure is defined as

$$s\equiv {\textstyle{1\over 2}}p,$$ (1)

where $p$ is the Perimeter. The semiperimeter of Polygons appears in unexpected ways in the computation of their Areas. The most notable cases are in the Altitude, Exradius, and Inradius of a Triangle, the Soddy Circles, Heron's Formula for the Area of a Triangle in terms of the legs $a$, $b$, and $c$

$$A_\Delta=\sqrt{s(s-a)(s-b)(s-c)},$$ (2)

and Brahmagupta's Formula for the Area of a Quadrilateral

$$A_{\rm quadrilateral}=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left({A+B\over 2}\right)}.$$ (3)

The semiperimeter also appears in the beautiful L'Huilier's Theorem about Spherical Triangles.

For a Triangle, the following identities hold,

$$s-a = {\textstyle{1\over 2}}(-a+b+c)$$ (4)

$$s-b = {\textstyle{1\over 2}}(a+b-c)$$ (5)

$$s-c = {\textstyle{1\over 2}}(a+b-c).$$ (6)

Now consider the above figure. Let $I$ be the Incenter of the Triangle $\Delta ABC$, with $D$, $E$, and $F$ the tangent points of the Incircle. Extend the line $BA$ with $GA=CE$. Note that the pairs of triangles $(ADI, AFI)$, $(BDI, BEI)$, $(CFI, CEI)$ are congruent. Then

$$BG = BD+AD+AG=BD+AD+CE$$

$$= {\textstyle{1\over 2}}(2BD+2AD+2CE)$$

$$= {\textstyle{1\over 2}}[(BD+BE)+(AD+AF)+(CE+CF)]$$

$$= {\textstyle{1\over 2}}[(BD+AD)+(BE+CE)+(AF+CF)]$$

$$= {\textstyle{1\over 2}}(AB+BC+AC)={\textstyle{1\over 2}}(a+b+c)=s.$$ (7)

Furthermore,

$$s-a = BG-BC$$

$$= (BD+AD+AG)-(BE+CE)$$

$$= (BD+AD+CE)-(BD+CE)=AC$$ (8)

$$s-b = BG-AC$$

$$= (BD+AD+AG)-(AF+CF)$$

$$= (BD+AD+CE)-(AD+CE)=BD$$ (9)

$$s-c = BG-AB=AG$$ (10)

(Dunham 1990). These equations are some of the building blocks of Heron's derivation of Heron's Formula.

References

Dunham, W. ``Heron's Formula for Triangular Area.'' Ch. 5 in Journey Through Genius: The Great Theorems of Mathematics. New York: Wiley, pp. 113-132, 1990.