Can anyone provide a geometric proof of $\frac1r = \frac{1}{h_a} + \frac{1}{h_b} + \frac{1}{h_c}$ in a given triangle, if there is one? Here, $h_i$ refers to the distance from the side $i$ to the opposite vertex, and $r$ is the inradius. I get the proof of the formula but I would like to get an intuitive understanding of it.
2026-02-24 05:37:27.1771911447
Is there a geometric proof of $\frac1r = \frac{1}{h_a} + \frac{1}{h_b} + \frac{1}{h_c}$ in a given triangle?
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Using $$2A=ah_a=bh_b=ch_c$$ and $$A=sr$$ we get $$\frac{s}{A}=\frac{a}{2A}+\frac{b}{2A}+\frac{c}{2A}$$ which is true, after cancelling the $$A$$ (area)