<s>[INST] Show that for every triangle $ABC$ $\frac{1}{h_b}+\frac{1}{h_c}-\frac{1}{h_a}=\frac{1}{r_a}$ Show that for every triangle $ABC$ $$\dfrac{1}{h_b}+\dfrac{1}{h_c}-\dfrac{1}{h_a}=\dfrac{1}{r_a}.$$
Let $S_{ABC}=S$. Then $$S=\dfrac{ah_a}{2}=\dfrac{bh_b}{2}=\dfrac{ch_c}{2}\Rightarrow h_a=\dfrac{2S}{a};h_b=\dfrac{2S}{b};h_c=\dfrac{2S}{c}.$$ On the other hand, $$S=r_a(p-a)\Rightarrow r_a=\dfrac{S}{p-a}.$$ where $r_a$ is the exradii.
So $$\dfrac{1}{h_b}+\dfrac{1}{h_c}-\dfrac{1}{h_a}=\dfrac{b+c-a}{2S}.$$ I can't see that this is equal to $\dfrac{1}{r_a}=\dfrac{p-a}{S}.$ Thank you in advance!
 [/INST] If $p$ is the semiperimeter $\frac {a+b+c}2$:
$$\frac1{h_b}+\frac1{h_c}-\frac1{h_a}=\frac {b+c-a}{2\color{red}S} = \frac {a+b+c-2a}{2\color{red}S} = \frac{\frac {a+b+c}2-\frac {2a}2}{\color{red}S} = \frac {p-a}{\color{red}S} = \frac1{r_a}$$
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