[Butterfly Theorem] It is known that circle O and PQ are chords, let m be the midpoint of chord PQ, and let m be chords AB and CD.

Let PQ where AD and BC intersect at points X and Y, then M is the midpoint of XY.

It is proved that the center O is perpendicular to AD and BC, and the vertical feet S and T connect OX, OY and OM. SM .MT .∫△SMD∽△CMB and SD= 1/2ADBT= 1/2BC,

∴DS/BT=DM/BM and ≈ D = ∠ B.

∴△MSD∽△MTB,∠MSD=∠MTB

∴∠msx=∠mty；

And o, s, x, m and o, t. Y .m are a four-point * * * circle,

∴∠XOM=∠YOM

∵OM⊥PQ

∴XM=YM

There is also an analytic geometry method, which is generalized.

[Summary] The three chords ab, CD and ef of the conic S intersect at a point M, ED and AB intersect at Q and Q, CF and AB intersect at P,

Then1/QM-1/pm =1/am-1/BM.

With m as the origin, AB as the x axis, s: ax 2+bxy+cy 2+dx+ey+f = 0,

CD:y=k 1x, EF:Y=k2x is a quadratic curve system equation passing through four points of c, d, e and f;

S+t(y-k 1x)(y-k2x)=0。

Let y=0 and get (a+TK 1K2) x 2+DX+F = 0, whose root is the abscissa of the intersection of the curve and the horizontal axis.

Then FX 2+DX+A+TK 1K2 = 0 is the reciprocal of the abscissa, and its sum =-D/F is a constant value.

That is1/QM+1/(-pm) =1/am+1/(-BM).

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