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Giả sử $h_1,h_2,h_3,h_4$ lần lượt là các chiều cao kẻ từ $A,B,C,D$
Gọi $I$ là tâm mặt cầu nội tiếp tứ diện.
Ta có:
$V_{IBCD}+V_{IACD}+V_{IABD}+V_{IABC}=V_{ABCD}$
$\Leftrightarrow \frac{V_{IBCD}}{V_{ABCD}}+\frac{V_{IACD}}{V_{ABCD}}+\frac{V_{IABD}}{V_{ABCD}}+\frac{V_{IABC}}{V_{ABCD}}=1$
$\Leftrightarrow \frac{\displaystyle\frac{1}{3}rS_{BCD}}{\displaystyle\frac{1}{3}h_1S_{BCD}}+\frac{\displaystyle\frac{1}{3}rS_{ACD}}{\displaystyle\frac{1}{3}h_2S_{ACD}}+\frac{\displaystyle\frac{1}{3}rS_{ABD}}{\displaystyle\frac{1}{3}h_3S_{ABD}}+\frac{\displaystyle\frac{1}{3}rS_{ABC}}{\displaystyle\frac{1}{3}h_4S_{ABC}}=1$
$\Leftrightarrow \frac{r}{h_1}+\frac{r}{h_2}+\frac{r}{h_3}+\frac{r}{h_4}=1$ , đpcm.
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