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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 $a(n_1+n_2+n_3+n_4)=2A$

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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.

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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). $\blacksquare$

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