반 아우벨의 정리(Van Aubel's Theorem)

.(Henricus Hubertus van Aubel)

삼각형 ABC (△ABC)의 각 꼭짓점에서 삼각형 ABC 내부의 임의의 한 점 P와 만나는 세개의 직선 AP, BP, CP가 각각 선분 BC, CA, AB와 만나는 점을 <math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle A_{1}}"><semantics><annotation encoding="application/x-tex">{\displaystyle A_{1}}</annotation></semantics></math>, <math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle B_{1}}"><semantics><annotation encoding="application/x-tex">{\displaystyle B_{1}}</annotation></semantics></math>, <math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle C_{1}}"><semantics><annotation encoding="application/x-tex">{\displaystyle C_{1}}</annotation></semantics></math>이라고 할 때 (<math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle A_{1}\in {\overline {BC}}}"><semantics><annotation encoding="application/x-tex">{\displaystyle A_{1}\in {\overline {BC}}}</annotation></semantics></math>, <math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle B_{1}\in {\overline {AC}}}"><semantics><annotation encoding="application/x-tex">{\displaystyle B_{1}\in {\overline {AC}}}</annotation></semantics></math>, <math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle C_{1}\in {\overline {AB}}}"><semantics><annotation encoding="application/x-tex">{\displaystyle C_{1}\in {\overline {AB}}}</annotation></semantics></math>)

다음과 같은 식이 성립하는데,

<math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle {\frac {AP}{PA_{1}}}={\frac {AC_{1}}{C_{1}B}}+{\frac {AB_{1}}{B_{1}C}}}"><semantics><annotation encoding="application/x-tex">{\displaystyle {\frac {AP}{PA_{1}}}={\frac {AC_{1}}{C_{1}B}}+{\frac {AB_{1}}{B_{1}C}}}</annotation></semantics></math>

이를 판 아우벌의 정리라고 한다.

증명 Ⅰ

<math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle P_{\triangle XYZ}}"><semantics><annotation encoding="application/x-tex">{\displaystyle P_{\triangle XYZ}}</annotation></semantics></math>는 삼각형 <math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle XYZ}"><semantics><annotation encoding="application/x-tex">{\displaystyle XYZ}</annotation></semantics></math>의 면적을 뜻한다.

삼각형 <math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle ABC}"><semantics><annotation encoding="application/x-tex">{\displaystyle ABC}</annotation></semantics></math>와 삼각형 <math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle PBC}"><semantics><annotation encoding="application/x-tex">{\displaystyle PBC}</annotation></semantics></math>는 동일한 선분 <math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle BC}"><semantics><annotation encoding="application/x-tex">{\displaystyle BC}</annotation></semantics></math>를 밑변으로 가지고 있으므로 높이(선분)의 비율은 면적의 비율과 동일하다. 따라서

<math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle {\frac {AA_{1}}{PA_{1}}}={\frac {P_{\triangle ABC}}{P_{\triangle BCP}}}}"><semantics><annotation encoding="application/x-tex">{\displaystyle {\frac {AA_{1}}{PA_{1}}}={\frac {P_{\triangle ABC}}{P_{\triangle BCP}}}}</annotation></semantics></math>가 성립하는데

이것은

<math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle {\frac {AP}{PA_{1}}}={\frac {P_{\triangle APC}+P_{\triangle APB}}{P_{\triangle BCP}}}}"><semantics><annotation encoding="application/x-tex">{\displaystyle {\frac {AP}{PA_{1}}}={\frac {P_{\triangle APC}+P_{\triangle APB}}{P_{\triangle BCP}}}}</annotation></semantics></math>을 내포한다.

또 삼각형 <math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle ACC_{1}}"><semantics><annotation encoding="application/x-tex">{\displaystyle ACC_{1}}</annotation></semantics></math>과 삼각형 <math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle BCC_{1}}"><semantics><annotation encoding="application/x-tex">{\displaystyle BCC_{1}}</annotation></semantics></math>을 보면 두 삼각형은 동일한 선분 <math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle CC_{1}}"><semantics><annotation encoding="application/x-tex">{\displaystyle CC_{1}}</annotation></semantics></math>을 같은 높이로 가지고 있으므로 밑변(선분)의 비율은 면적의 비율과 동일하다. 따라서

<math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle {\frac {AC_{1}}{C_{1}B}}={\frac {P_{\triangle ACC_{1}}}{P_{\triangle BCC_{1}}}}}"><semantics><annotation encoding="application/x-tex">{\displaystyle {\frac {AC_{1}}{C_{1}B}}={\frac {P_{\triangle ACC_{1}}}{P_{\triangle BCC_{1}}}}}</annotation></semantics></math>가 성립한다.

위와 동일한 방법으로

<math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle {\frac {AC_{1}}{C_{1}B}}={\frac {P_{\triangle AC_{1}P}}{P_{\triangle BC_{1}P}}}}"><semantics><annotation encoding="application/x-tex">{\displaystyle {\frac {AC_{1}}{C_{1}B}}={\frac {P_{\triangle AC_{1}P}}{P_{\triangle BC_{1}P}}}}</annotation></semantics></math>을 얻을 수 있다.

따라서

<math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle {\frac {AC_{1}}{C_{1}B}}={\frac {P_{\triangle ACP}+P_{\triangle AC_{1}P}}{P_{\triangle BCP}+P_{\triangle BC_{1}P}}}={\frac {P_{\triangle AC_{1}P}}{P_{\triangle BC_{1}P}}}}"><semantics><annotation encoding="application/x-tex">{\displaystyle {\frac {AC_{1}}{C_{1}B}}={\frac {P_{\triangle ACP}+P_{\triangle AC_{1}P}}{P_{\triangle BCP}+P_{\triangle BC_{1}P}}}={\frac {P_{\triangle AC_{1}P}}{P_{\triangle BC_{1}P}}}}</annotation></semantics></math>

위 식을 정리하면,

(i)   <math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle {\frac {AC_{1}}{C_{1}B}}={\frac {P_{\triangle ACP}}{P_{\triangle BCP}}}}"><semantics><annotation encoding="application/x-tex">{\displaystyle {\frac {AC_{1}}{C_{1}B}}={\frac {P_{\triangle ACP}}{P_{\triangle BCP}}}}</annotation></semantics></math>

이와 같은 방법으로 동일하게 구하면

(ii)  <math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle {\frac {AB_{1}}{B_{1}C}}={\frac {P_{\triangle APB}}{P_{\triangle BCP}}}}"><semantics><annotation encoding="application/x-tex">{\displaystyle {\frac {AB_{1}}{B_{1}C}}={\frac {P_{\triangle APB}}{P_{\triangle BCP}}}}</annotation></semantics></math>

(i)식과 (ii)식을 각각 더하면

<math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle {\frac {AB_{1}}{B_{1}C}}+{\frac {AC_{1}}{C_{1}B}}={\frac {P_{\triangle APC}+P_{\triangle APB}}{P_{\triangle BCP}}}={\frac {AP}{PA_{1}}}}"><semantics><annotation encoding="application/x-tex">{\displaystyle {\frac {AB_{1}}{B_{1}C}}+{\frac {AC_{1}}{C_{1}B}}={\frac {P_{\triangle APC}+P_{\triangle APB}}{P_{\triangle BCP}}}={\frac {AP}{PA_{1}}}}</annotation></semantics></math>

판 아우벌의 정리를 증명할 수 있다.

증명 Ⅱ

판 아우벌의 정리는 메넬라오스의 정리와 체바의 정리를 이용하여 증명할 수 있다.

삼각형 <math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle ABA_{1}}"><semantics><annotation encoding="application/x-tex">{\displaystyle ABA_{1}}</annotation></semantics></math>과 직선 <math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle CC_{1}}"><semantics><annotation encoding="application/x-tex">{\displaystyle CC_{1}}</annotation></semantics></math>에 대해서 메넬라오스의 정리가 성립한다.

<math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle {\frac {AC_{1}}{C_{1}B}}\cdot {\frac {BC}{CA_{1}}}\cdot {\frac {A_{1}P}{PA}}=1}"><semantics><annotation encoding="application/x-tex">{\displaystyle {\frac {AC_{1}}{C_{1}B}}\cdot {\frac {BC}{CA_{1}}}\cdot {\frac {A_{1}P}{PA}}=1}</annotation></semantics></math>

이 식을 변형하면,

<math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle {\frac {PA}{A_{1}P}}={\frac {AC_{1}}{C_{1}B}}\cdot {\frac {BC}{CA_{1}}}={\frac {AC_{1}}{C_{1}B}}\cdot {\frac {BA_{1}+CA_{1}}{CA_{1}}}={\frac {AC_{1}}{C_{1}B}}\cdot ({\frac {BA_{1}}{CA_{1}}}+{\frac {CA_{1}}{CA_{1}}})}"><semantics><annotation encoding="application/x-tex">{\displaystyle {\frac {PA}{A_{1}P}}={\frac {AC_{1}}{C_{1}B}}\cdot {\frac {BC}{CA_{1}}}={\frac {AC_{1}}{C_{1}B}}\cdot {\frac {BA_{1}+CA_{1}}{CA_{1}}}={\frac {AC_{1}}{C_{1}B}}\cdot ({\frac {BA_{1}}{CA_{1}}}+{\frac {CA_{1}}{CA_{1}}})}</annotation></semantics></math>

따라서,

(i) <math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle {\frac {PA}{A_{1}P}}={\frac {AC_{1}}{C_{1}B}}\cdot {\frac {BA_{1}}{CA_{1}}}+{\frac {AC_{1}}{C_{1}B}}}"><semantics><annotation encoding="application/x-tex">{\displaystyle {\frac {PA}{A_{1}P}}={\frac {AC_{1}}{C_{1}B}}\cdot {\frac {BA_{1}}{CA_{1}}}+{\frac {AC_{1}}{C_{1}B}}}</annotation></semantics></math>

삼각형 <math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle ABC}"><semantics><annotation encoding="application/x-tex">{\displaystyle ABC}</annotation></semantics></math>에서 체바의 정리가 성립한다.

<math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle {\frac {AC_{1}}{C_{1}B}}\cdot {\frac {BA_{1}}{A_{1}C}}\cdot {\frac {CB_{1}}{B_{1}A}}=1}"><semantics><annotation encoding="application/x-tex">{\displaystyle {\frac {AC_{1}}{C_{1}B}}\cdot {\frac {BA_{1}}{A_{1}C}}\cdot {\frac {CB_{1}}{B_{1}A}}=1}</annotation></semantics></math>

따라서,

(ii) <math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle {\frac {B_{1}A}{CB_{1}}}={\frac {AC_{1}}{C_{1}B}}\cdot {\frac {BA_{1}}{A_{1}C}}}"><semantics><annotation encoding="application/x-tex">{\displaystyle {\frac {B_{1}A}{CB_{1}}}={\frac {AC_{1}}{C_{1}B}}\cdot {\frac {BA_{1}}{A_{1}C}}}</annotation></semantics></math>

(ii)식을 (i)식에 대입하면

<math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle {\frac {PA}{A_{1}P}}={\frac {B_{1}A}{CB_{1}}}+{\frac {AC_{1}}{C_{1}B}}}"><semantics><annotation encoding="application/x-tex">{\displaystyle {\frac {PA}{A_{1}P}}={\frac {B_{1}A}{CB_{1}}}+{\frac {AC_{1}}{C_{1}B}}}</annotation></semantics></math>