let us assume that the order is as follows $a,b,c,2*r$ then we can easily conclude that $r=\frac{b+c-a}{2}$ if we take any other order then either $r=\frac{c+a-b}{2}$ or $r=\frac{a+b-c}{2}$ let us assume that $r=\frac{b+c-a}{2}=(s-a)$ we know that $r=(s-a)tan\bigg(\frac{A}{2}\bigg)$ this shows $tan\bigg(\frac{A}{2}\bigg)=1$ thus $\angle A=90$ if we assume the other relations to be true then we either get $\angle B=90$ or $\angle C=90$ Hence proved