دایره محاطی خارجی

اثبات

\(\begin{array}{l}{S_{A\mathop B\limits^\Delta C}} = S = {S_{O\mathop A\limits^\Delta B}} + {S_{O\mathop A\limits^\Delta C}} - {S_{O\mathop B\limits^\Delta C}}\\\\ \Rightarrow \left( {\frac{1}{2}{r_a} \times AB} \right) + \left( {\frac{1}{2}{r_a} \times AC} \right) - \left( {\frac{1}{2}{r_a} \times BC} \right)\\\\ \Rightarrow S = {r_a}\left( {\frac{{AB + AC - BC}}{2}} \right)\\\\ \Rightarrow {r_a}\left( {\frac{{AB + AC + BC - BC - BC}}{2}} \right)\\\\ \Rightarrow {r_a}\left( {\frac{{2P - 2a}}{2}} \right) \Rightarrow {r_a} = \frac{S}{{P - a}}\end{array}\)

با توجه به اینکه \({r_a} = \frac{S}{{P - a}}\) ، بنابراین:

\(\begin{array}{l}\frac{1}{{{r_a}}} + \frac{1}{{{r_b}}} + \frac{1}{{{r_c}}} = \frac{{P - a}}{S} + \frac{{P - b}}{S} + \frac{{P - c}}{S}\\\\ \Rightarrow \frac{{3P - \left( {a + b + c} \right)}}{S} = \frac{{3P - 2P}}{S} = \frac{P}{S} = \frac{1}{r}\end{array}\)

می دانیم مساحت مثلث ABC با توجه به ارتفاع های سه ضلع آن به صورت زیر بدست می آید:

\(\begin{array}{l}{S_{A\mathop B\limits^\Delta C}} = \frac{1}{2}{h_a} \times a = \frac{1}{2}{h_b} \times b = \frac{1}{2}{h_c} \times c\\\\ \Rightarrow 2S = {h_a} \times a = {h_b} \times b = {h_c} \times c\\\\ \Rightarrow {h_a} = \frac{{2S}}{a}\;,\;{h_b} = \frac{{2S}}{b}\;,\;{h_c} = \frac{{2S}}{c}\\\\ \Rightarrow \frac{1}{{{h_a}}} + \frac{1}{{{h_b}}} + \frac{1}{{{h_c}}} = \frac{1}{{\frac{{2S}}{a}}} + \frac{1}{{\frac{{2S}}{b}}} + \frac{1}{{\frac{{2S}}{c}}}\\\\ \Rightarrow \frac{a}{{2S}} + \frac{b}{{2S}} + \frac{c}{{2S}} = \frac{{a + b + c}}{{2S}} = \frac{{2P}}{{2S}} = \frac{1}{r}\end{array}\)