Incircle and Inradius

10.7 Incircle and Inradius

Theorem 1. $r=\dfrac{\Delta}{s}$.

Proof. The incentre $I$ is equidistant (distance $r$) from all three sides. Joining $I$ to $A,B,C$ splits $\triangle ABC$ into three triangles $IBC,ICA,IAB$ with heights $r$ on bases $a,b,c$ respectively. So $\Delta=\frac12ar+\frac12br+\frac12cr=\frac r2(a+b+c)=rs$, giving $r=\Delta/s$.

Theorem 2. $r=(s-a)\tan\dfrac A2$.

Proof. The tangent length from vertex $A$ to the incircle is $s-a$ (a standard tangent-length result: if the incircle touches $AB$ at $P$ and $AC$ at $Q$, then $AP=AQ=s-a$). In the right triangle formed by $A$, the touch point $P$, and the incentre $I$, the angle at $A$ is $\frac A2$ (since $AI$ bisects angle $A$), the side adjacent to this angle is $AP=s-a$, and the side opposite is $IP=r$. Hence $\tan\frac A2=\dfrac{r}{s-a}$, i.e. $r=(s-a)\tan\frac A2$.

Theorem 3. $r=4R\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2$.

Proof (sketch). Combine $r=(s-a)\tan\frac A2$ with the half-angle formula $\tan\frac A2=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ from §10.5: $r=(s-a)\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}=\frac{\sqrt{s(s-a)(s-b)(s-c)}}{s}=\frac\Delta s$, recovering Theorem 1 and confirming consistency. Writing $\Delta=2R^2\sin A\sin B\sin C$ (§10.6) and each $\sin A=2\sin\frac A2\cos\frac A2$, then dividing by $s=4R\cos\frac A2\cos\frac B2\cos\frac C2$ (itself obtainable from the projection-rule sum of §10.3 written in half-angle form) gives $r=4R\sin\frac A2\sin\frac B2\sin\frac C2$ directly.

Worked example. For $a=13,b=14,c=15$: $s=21,\Delta=84$, so $r=\Delta/s=84/21=4$. This ties directly into Exercise 10(b), Q1.

A worked example combining SAS data. For $A=60^\circ,b=8,c=10$ (Exercise 10(b), Q3): first $\Delta=\frac12bc\sin A=\frac12(8)(10)\sin60^\circ=20\sqrt3$, and $a^2=b^2+c^2-2bc\cos A=84\Rightarrow a=2\sqrt{21}$, giving $s=9+\sqrt{21}$. Then $r=\Delta/s=\dfrac{20\sqrt3}{9+\sqrt{21}}$, which rationalises to $3\sqrt3-\sqrt7\approx2.550$ — showing that even when the semi-perimeter is irrational, the inradius formula $r=\Delta/s$ still applies directly; it is simply a matter of careful rationalisation at the end. …