Non-circular form of butterfly theorem;
As shown in the figure, the two chords AB and CD in the extended circle O intersect at the point M, the PM is the OM vertical line, and the vertical line intersects with the extended lines CB and AD at E and F, so that ME=MF can be obtained (see Butterfly Theorem Proof 2, 3 and 4 for the proof method).
1. In the ellipse
As shown in figure 1, the major axes A 1 and A2 of the ellipse are parallel to the X axis, and the minor axis B 1B2 is on the Y axis.
The center is M(o, r) (b >; r & gt0)。
(1) Write the equation of ellipse, and find the focal coordinates and eccentricity of ellipse.
(II) The straight line y=k 1x intersects the ellipse at two points C(x 1, y 1) and D(x2, y2) (y2 >; 0); The straight line y=k2x intersects the ellipse at two points G(x3, y3), H(x4, y4) (y4 >; 0)。 Verification: k1x1x 2/(x1+x2) = k2x3x4/(x3+x4)
(III) For C, D, G and H in (II), let CH pass through the X axis at point P and GD pass through the X axis at point Q. ..
Proof: | OP | = | OQ |. (The proof process does not consider the case that CH or GD is perpendicular to the X axis. )
Take AM and DM as vertical lines from X, and let the vertical feet be X' and X'' respectively.
Similarly, BM and CM are taken as vertical lines from y, and the vertical feet are y' and y' respectively.
Let: k 1 x1x 2/(x1+x2) = k2x3x4/(x3+x4) be the formula (1), and the two sides are reciprocal, thus obtaining
1/k 1x 2+ 1/k 1x 1 = 1/k2 x4+ 1/k2x 3①'
Let x1x 4/(k1x1-k2x4) =-x2x3/(k1x2-k2x3) be of type ②, and the two sides are reciprocal to obtain k1/x4-k2/.
①' Multiply both sides by k 1 k2 to get the product.
k2/x 1+k2/x2 = k 1/x3+k 1/x4
It is exactly the same as ②'. Here, it is easier to achieve the goal by using the method of simultaneous deformation of two formulas, with analysis, synthesis, thinking and operation. The choice of thinking depends on the observation and association of the characteristics of the formula.
Looking at the characteristics and solving process of this problem, we see the function and power of using algebraic equation method to deal with geometric problems.
2. In the conical section
Through projective geometry, we can easily extend the butterfly theorem to ordinary arbitrary conic curves (including ellipses, hyperbolas, parabolas, and even degenerate into two intersecting straight lines).
The midpoint of PQ on the chord of conic curve C is m, the intersection point M is defined as two chords AB and CD, and chords AD and BC intersect PQ at X and Y respectively, so M is the midpoint of XY.
This theorem can be easily proved by projection transformation.
There is an important conclusion about projective transformation in projective geometry. For any two quadratic curves C 1 and C2 on the plane, there is a unique projection transformation to transform curve C65438 by arbitrarily specifying points A 1 inside C 1 and points B 1 above C 1 and arbitrarily specifying points A2 inside C2 and B2 above C2.
So for this problem, we can transform C 1 into a circle m by projection transformation, and transform the middle point m of the chord PQ into the center of the circle.
After this transformation, chords AB and CD are the diameters of circle M, quadrilateral ACBD is the inscribed rectangle of circle M, and PQ is also the diameter. Obviously, m is the midpoint of x and y after projection transformation with symmetry. Moreover, because M is the midpoint of the straight line PQ before and after the transformation, it can be concluded that this transformation is an affine transformation on the straight line PQ, so M before the transformation is also the midpoint of XY.
3. In the parallelogram
In parallelogram, m is the midpoint of diagonal AB and CD.
4. Candy Theorem
If the midpoint condition is removed, the conclusion becomes a general proportional formula about vectors, which is called "candy theorem", and both 2 and 3 hold.