Ta có:

$S = \frac{1}{2}a.{h_a} \Rightarrow \frac{1}{{{h_a}}} = \frac{a}{{2S}}$;

$S = \frac{1}{2}b.{h_b} \Rightarrow \frac{1}{{{h_b}}} = \frac{b}{{2S}}$;

$S = \frac{1}{2}c.{h_c} \Rightarrow \frac{1}{{{h_c}}} = \frac{c}{{2S}}$.

Do đó: $\frac{1}{{{h_a}}} + \frac{1}{{{h_b}}} + \frac{1}{{{h_c}}} = \frac{{a + b + c}}{{2S}} = \frac{{2p}}{{2S}} = \frac{p}{S} = \frac{p}{{p.r}} = \frac{1}{r}$.

Hay $\frac{1}{{{h_a}}} + \frac{1}{{{h_b}}} + \frac{1}{{{h_c}}} = \frac{1}{r}$ (đpcm).