Mathematics - Properties of Triangle Question with Solution | TestHub

We know the formulas for inradius and exradii in terms of area $\Delta$ and semi-perimeter $s$: $r=\frac{\Delta }{s}$, $r_1=\frac{\Delta }{s-a}$, $r_2=\frac{\Delta }{s-b}$, $r_3=\frac{\Delta }{s-c}$. Substituting these into the given relation $r_1=r+r_2+r_3$: $\frac{\Delta }{s-a}=\frac{\Delta }{s}+\frac{\Delta }{s-b}+\frac{\Delta }{s-c}$. Dividing by $\Delta$ (since $\Delta \ne 0$): $\frac{1}{s-a}=\frac{1}{s}+\frac{1}{s-b}+\frac{1}{s-c}$. This identity holds true if and only if the triangle is right-angled at A. For example, if $A=90^{\circ }$, then $r_1=s\tan (A/2)=s\tan (45^{\circ })=s$, and $r=s\tan (A/2)\tan (B/2)\tan (C/2)=(s-a)\tan (A/2)=s-a$. Also $r_2=s\tan (B/2)$ and $r_3=s\tan (C/2)$. A known identity is $r_1=4R\sin (A/2)\cos (B/2)\cos (C/2)$, $r=4R\sin (A/2)\sin (B/2)\sin (C/2)$, etc. The given relation $r_1=r+r_2+r_3$ simplifies to $\cot (A/2)=\cot (B/2)+\cot (C/2)-\cot (A/2)$. This is equivalent to $\tan (A/2)=\tan (B/2)+\tan (C/2)-\tan (A/2)$, which is true if $A=90^{\circ }$.