STATEMENT 1: In a ΔABC,(1−r1r2)(1−r1r3)=2 then the triangle is right angled.STATEMENT 2: In a ΔABC, r1r2+r2r3+r3r1=2r2
(1−r1r2)(1−r1r3)=2(1−s−bs−a)(1−s−cs−a)=2(s−a−s+bs−a)(s−a−s+cs−a)=2(b−a)(c−a)=2(s−a)(s−a)