# 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

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, $\alpha = \beta$ $\Rightarrow \space AC||BD$ $\Rightarrow \space [\Delta BDA]=[\Delta BDC]$ $\Rightarrow \space [\Delta AIB]=[\Delta CID]$.