MarkBcc168 wrote: Suppose that $f : \mathbb{R}^+\to\mathbb R$ satisfies the equation $$f(a+b+c+d) = f(a)+f(b)+f(c)+f(d)$$for all $a,b,c,d$ that are the four sides of some tangential quadrilateral. Show that $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}^+$. Choosing a square with side $a$, we have $f(4a)=4f(a)$ $\forall a>0$ Choosing a triangle $ABC$ with sides $AB=a$, $BC=b$, angle ABC greater than $\frac{\pi}2$ and point D symetrical of B regarding AC, we have : $f(2a+2b)=2f(a)+2f(b)$ $\forall a,b>0$ So $2f(2a+2b)=f(4a)+f(4b)$ and so $f(x)+f(y)=2f(\frac{x+y}2)$ $\forall x,y>0$ So $f(x)+f(y+z)=f(x+y)+f(z)$ ($=2f(\frac{x+y+z}2)$) $\forall x,y,z>0$ So $f(x+y)-f(x)=f(z+y)-f(z)$ depends only on $y$ and is $h(y)$ for some function $h$ So $f(x+y)=f(x)+h(y)$ So $f(x)+h(y)=f(y)+h(x)$ and $h(x)-f(x)$ is some constant $c$ And $f(x+y)=f(x)+f(y)+c$ $\forall x,y>0$ And $f(4x)=4f(x)$ implies $c=0$ And so $\boxed{f(x+y)=f(x)+f(y)\quad\forall x,y>0}$, as required