Areal Butterflies

Sidney Kung
May 14, 2012

Circles \(O_1\) and \(O_2\) intersect in points \(M\) and \(N\). Line passing through \(M\) intersect\(O_1\) and \(O_2\) in \(A\) and \(B\), respectively. Line passing through \(N\) intersect\(O_1\) and \(O_2\) in \(C\) and \(D\), respectively. If \(AB\) does not intersect \(CD\), and if \(AD\cap BC=I\), then \([\Delta AIB]=[\Delta CID]\), where, \([\Omega ]\) indicates the area of figure \(\Omega\).

Proof

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Copyright © 1996-2018 Alexander Bogomolny

Circles \(O_1\) and \(O_2\) intersect in points \(M\) and \(N\). Line passing through \(M\) intersect\(O_1\) and \(O_2\) in \(A\) and \(B\), respectively. Line passing through \(N\) intersect\(O_1\) and \(O_2\) in \(C\) and \(D\), respectively. If \(AB\) does not intersect \(CD\), and if \(AD\cap BC=I\), then \([\Delta AIB]=[\Delta CID]\), where, \([\Omega ]\) indicates the area of figure \(\Omega\).

Proof

There are two cases to consider:

  \(A,M,N,C\in O_{1},\space \alpha +\gamma = \pi\)
\(B,M,N,D \in O_{2},\space \gamma = \beta\)
Therefore, \(\alpha + \beta = \pi\)
\(\Rightarrow \space AC||BD\)
\(\Rightarrow \space [\Delta ACB]=[\Delta CAD]\)
\(\Rightarrow \space [\Delta AIB]=[\Delta CID]\).
  \(A,M,N,C\in O_{1},\space \alpha = \gamma\)
\(B,M,N,D \in O_{2},\space \gamma = \beta\)
Therefore, \(\alpha = \beta\)
\(\Rightarrow \space AC||BD\)
\(\Rightarrow \space [\Delta BDA]=[\Delta BDC]\)
\(\Rightarrow \space [\Delta AIB]=[\Delta CID]\).
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|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny
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