To prove:
(r1−r)(r2+r3)=a2We know,
r1=4RsinA2cosB2cosC2r2=4RsinB2cosC2cosA2r3=4RsinC2cosA2cosB2r=4RsinA2sinB2sinC2
Now,
(r1−r)(r2+r3)=(4RsinA2cosB2cosC2−4RsinA2sinB2sinC2)(4RsinB2cosC2cosA2+4RsinC2cosA2cosB2)=(4RsinA2(cosB2cosC2−sinB2sinC2))(4RcosA2(sinB2cosC2+sinC2cosB2))=(4RsinA2(cosB+C2))(4RcosA2(sinB+C2))=4Rsin2A2cos2A2⋅4R2(2sinA2cosA2)2=4R2sin2A=(2RsinA)2=a2