# Formula for the area of a rhombus

Can you prove the following claim? The claim is inspired by Harcourt's theorem.

In any rhombus $$ABCD$$ construct an arbitrary tangent to the incircle of rhombus . Let $$n_1,n_2,n_3,n_4$$ be a signed distances from vertices $$A,B,C,D$$ to tangent line respectively, such that distances to a tangent from points on opposite sides are opposite in sign, while those from points on the same side have the same sign. Denote the side length of rhombus by $$a$$ and the area of rhombus by $$A$$ , then $$a(n_1+n_2+n_3+n_4)=2A$$

GeoGebra applet that demonstrates this claim can be found here. I tried to adapt the proof of Harcourt's theorem given in this paper but without success.

Let $$r$$ be the inradius of the rhombus. Clearly, since $$A$$ and $$C$$ are symmetric about $$I$$, as are $$B$$ and $$D$$, we have$$(n_1-r)+(n_3-r)=0,$$ $$(n_2-r)+(n_4-r)=0.$$ Hence, it suffices to prove $$2ar=A.$$ However, this is an immediate consequence of the fact that any polygon's area is equal to its semiperimeter times its inradius (whenever the latter exists). $$\blacksquare$$