Butterfly theorem

Statement

Let $PQ$ be a chord of a circle in the Euclidean plane and let $M$ be its midpoint. Let $AB$ and $CD$ be two further chords through $M$ . Suppose that the segment $AD$ meets $PQ$ at $X$ and the segment $BC$ meets $PQ$ at $Y$ . Then $M$ is the midpoint of $XY$ .

Remarks

One proof drops perpendiculars from $X$ and $Y$ to the chords and uses similar triangles together with the power of the point $M$ , $MA \cdot MB = MC \cdot MD = MP \cdot MQ$ . A projective proof uses the involution that the pencil of conics through $A, B, C, D$ induces on the line $PQ$ . Example: let $CD$ be the mirror image of $AB$ in the diameter through $M$ perpendicular to $PQ$ , with $C$ the image of $A$ . Then the figure is symmetric and $MX = MY$ is clear. The theorem stays true when the circle is replaced by any conic.

Reading path

To understand this page from the ground up, read these 3 pages in order. The foundations come first.

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Each page uses the pages above it. The colour of the bar on a box shows how strong the evidence is, and the box also gives it as text. The list has the same pages.

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Circle concept
Midpoint concept
Butterfly theorem statement

Tag	`E9TX1NDAHB`
URI	`m2://E9TX1NDAHB`
Kind	statement, theorem
Writing	W0 stub
MSC	51M04
Status	This copy of the site has no build data. Run `m2 status E9TX1NDAHB`.
About	Circle
Uses	Circle, Midpoint
Same as	Wikidata Q377047 Wikipedia Butterfly theorem

Butterfly theorem

Statement

Remarks

Reading path

Further reading

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What links here