Clearly,

${\displaystyle \Delta =\frac{1}{2}r.\mathrm{AB}+\frac{1}{2}r.\mathrm{BC}+\frac{1}{2}r.\mathrm{CA},}$

so that

${\displaystyle \frac{1}{r}=\frac{\mathrm{AB}+\mathrm{BC}+\mathrm{CA}}{2\Delta }.}$

Also,

${\displaystyle \Delta =\frac{1}{2}{h}_1.\mathrm{BC}=\frac{1}{2}{h}_2,\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\mathrm{CA}=\frac{1}{2}{h}_3.\mathrm{AB},}$

thus

${\displaystyle \frac{1}{h_1}+\frac{1}{h_2}+\frac{1}{h_3}=\frac{\mathrm{BC}+\mathrm{CA}+\mathrm{AB}}{2\Delta }=\frac{1}{r}.}$

Now let $A\text{'}$, $B\text{'}$ and $C\text{'}$ be the centres of the escribed circles; see the diagram below. Also, for any triangle with vertices $U$, $V$ and $W,$ let $\Delta (U,V,W)$ denote its area. "

Considering the area of the kite $\mathrm{ABA}\text{'}C$ by splitting it into two triangles in two different ways we get

${\displaystyle \Delta (A,B,C)+\Delta (B,A\text{'},C)=\mathrm{area}(\mathrm{ABA}\text{'}C)=\Delta (A,B,A\text{'})+\Delta (A,C,A\text{'}).}$

This gives (with $\Delta =\Delta (A,B,C))$

${\displaystyle \Delta +\frac{1}{2}{r}_1.\mathrm{BC}=\frac{1}{2}{r}_1.\mathrm{AB}+\frac{1}{2}{r}_1.\mathrm{CA},}$

and hence

${\displaystyle \frac{2\Delta }{r_1}=\mathrm{AB}+\mathrm{CA}-\mathrm{BC}.}$

Similarly,

${\displaystyle \frac{2\Delta }{r_2}=\mathrm{AB}+\mathrm{BC}-\mathrm{CA},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\frac{2\Delta }{r_3}=\mathrm{CA}+\mathrm{BC}-\mathrm{AB},}$

and adding these we get

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{2\Delta (\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_C})}& =(\mathrm{AB}+\mathrm{CA}-\mathrm{BC})+(\mathrm{AB}+\mathrm{BC}-\mathrm{CA})+(\mathrm{CA}+\mathrm{BC}-\mathrm{AB})\\ \multicolumn{1}{c}{}& =(\mathrm{AB}+\mathrm{BC}+\mathrm{CA})\\ \multicolumn{1}{c}{}& =\frac{2\Delta }{r}\end{array}}$

as required.