# Butterfly in Inscriptible Quadrilateral

Here's a problem by Dao Thanh Oai with a solution by Telv Cohl.

Let $ABC$ be an inscriptible quadrilateral with the incircle $(O),$ $E=AC \cap BD.$ Let a line through $E$ meets $AD,$ $BC$ in $F$ and $G,$ respectively. Assume also, it meets $(O)$ at $H$ and $I,$ as in the diagram.

Prove that $EI=EH$ iff $EF=EG.$

### Proof

If $(O)$ touches $AD,$ $BC$ at $X,$ $Y,$ respectively, then, as is well known, $X,$ $E,$ $Y$ are collinear. Desargues' Involution Theorem applied to the degenerate quadrangle $XXYY,$ informs us that $E$ is a double fixed point of the involution defined on the given line, with $F,G$ and $H,I$ reciprocal pairs.

The condition $EI=EH$ means that the involution is a symmetry in $E,$ implying $EF=EG.$

**Note**: the proof above is reminiscent of Hubert Shutrick's proof (Proof #20) of the common Butterfly Theorem.

### Reference

- Michael Woltermann, Desargues’ Involution Theorem

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