(Source: Olympiad Series Second Edition High School Volume Plane Geometry Fan Duan Xi Deng Bowen Inverse and Polar Coordinate Matching P095 Case 9)

As shown in the figure, it is known that in a circle, it is the diameter and the intersection point is a straight line. Cut in a straight line (both are outside the circle). Connect,. Intersect at one point. Through, make the tangent of the circle respectively. Intersect at one point and connect. Verification: perpendicular to.

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If parallel lines intersect and tangents intersect, the polar line is on the tangent and still on the polar line, so the polar line passes through,,, and forms a harmonic point series, so,, and forms a harmonic beam, so it is a harmonic quadrilateral.

So it is tangent, that is, on the polar line, then the polar line passes through this point, and the combination is vertical, that is, vertical.

202 1- 10-3 1-02

(Source: Duan Xi Deng Bowen's Inverse and Pole Matching P095 Case 10).

Given a circle,,,, suppose and, and intersect at points and, and, and respectively. If, and four-point * * * circle (or * * * line), prove:, and four-point * * circle (or * * line).

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For an inverse transformation with an inverse center, then the inverse of is the inverse of the circumscribed circle,,, and of a straight line, and these two circles intersect.

As shown in the figure, as long as the certificate,, and four points are * * * round.

This turns into the problem of Mick points in triangles:

China, China and China are on the edge, China and China, respectively. If the circumscribed circle of China intersects the circumscribed circle of China, the circumscribed circle of China also crosses this point.

202 1- 10-3 1-03

(Source: Mathematical Olympiad Series, 2nd Edition, Plane Geometry of Senior High School Volume, Fan, Deng Bowen, Inverse and Polar Matching, P095 cases, 1 1)

As shown in the figure, the convex quadrilateral has an inscribed circle, and the inscribed circle cuts the edges,,,,, points,,, respectively, which are the midpoints of the line segments,,,, respectively. It is proved that a quadrilateral is a rectangle if and only if it is a circle.

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If the inscribed circle is taken as the inversion circle, then according to inversion theorem 4, the inverse points of, and are respectively, and because the inverse of the inversion center circle is still a circle, the * * * circle is equivalent to, * * * circle.

Note that,,, and are the midpoint of the quadrilateral, so the quadrilateral is a parallelogram. Therefore, the necessary and sufficient condition of the four-point * * * circle of,, and is that the parallelogram is a rectangle, which is equivalent to the,, * * * circle.

202 1- 10-3 1-04

(Source: Duan Xi Deng Bowen's Inverse and Pole Matching P096 Case 12).

A quadrilateral inscribed in a circle has a little satisfaction. Projected on the floor is. Prove:

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Proof 1: Because you only need proof.

Say it again,

So you only need a certificate (1).

Because of this, the circumscribed circle and the circumscribed circle are tangent to the point.

It is thought that to find the inverse center, the power of the circumscribed circle is the inverse power, and,, respectively become,, and are similar to the intersection of the circumscribed circle,, and.

Because the circumscribed circle is made up of.

Therefore, it is an isosceles trapezoid according to the inversion property.

Inverse evolution formula of slave distance

, (is the inverse power).

Therefore, (1), this formula is proved to be true, so it is original.

Proof 2: work, where is the point, submitted to.

Yuki。

It is easy to know the tangent of the circumscribed circle.

From the theorem of root axis, we know that we intersect at one point and let be.

From the known conditions, it is easy to know that there are four * * * cycles.

Therefore,

therefore

202 1- 10-3 1-05

(Source: Duan Xi Deng Bowen's Inverse and Pole Matching P097 Case 13).

Two-center quadrilateral with inner and outer centers respectively. Verification:,, three-point * * line.

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Prove a lemma first.

Lemma: A circle circumscribes a quadrilateral, and the tangent points are four lines * * *.

Proof of lemma: as shown in the figure, suppose that, is obtained by sine theorem.

In the same way.

So, that's.

So, the third line * * * points.

Similarly, point * * * in the third line is proved by lemma.

Back to the original question: as shown in the figure, the tangent point is still recorded as,, and by lemma.

Take the center of the circle as a circle, and, and are the midpoint of the four sides respectively.

From, it is called a parallelogram.

And,,, * * * circle, * * * circle, must be a rectangle, its center is set to, and there is.

Know the three-point * * * line from the inversion property.

If the midpoint is, then

According to the vertical diameter theorem, it is called a rectangle.

Therefore.

Therefore, it is the three-point * * line, thus, the three-point * * * line.