High School Math Contest Plane Geometry, a Contest Plane Geometry?

As shown in the figure: let the center of the circle be the coordinate point and the radius is 2 [this is to calculate the coordinates of e, b, f, h, and then find the equation of a circle according to the coordinates of bfh, if it can be done, you can prove that the three-point bfh is coherent, and then bring the coordinates of e into the same equation, if it conforms to the relation, then the four-point ebfh four-point contour] So e(o,o) b(root number 2, root number 2) f(-root number 2, 3 root number 2) [The coordinates of f can first find the intersection point P of the y-axis and bf, according to p, b two points to find the straight line, so as to get the y value of f. See below for details:

b(root number 2, root number 2) p(0,2 root number 2) according to the two-point formula to find the linear equation: (x-x1) (x2-x1) = (y-y1) (y2-y1) to obtain the linear equation: root number 2x + root number 2y = 4 and then take the x value of f = - root number 2 (the value of f x is the same as the value of x of a is - root number 2) to obtain the y value of f 3 root number 2

Then find the coordinates of h [according to the triangle HPE is an isosceles right triangle] h (-2 root number 2, 2 root number 2).

Let the equation for the circle of ebfh be x 2+y 2+dx+ey+f=0 and bring the coordinates of the three points of the point bfh into the equation x 2+y 2+dx+ey+f=0

Find d = root number 2 e = -3 root number 2 f = 0 So the equation for the circle of EBHF is x 2 + y 2 + root number 2x - 3 root number 2y = 0

Then bring the equation of the point e(0,0) into the equation that satisfies the circle. So EBFH is a four-point circle!

Proof of the symmetry point f with respect to the straight line ac point f', easy to know point f'On ag, fea= f'ea

bf //dg

fa ad=ba ag, i.e. fa ga=ab ad=ab ghc= gbc

g, h, b, c are four points in a circle.

It is necessary to prove that b, h, f, and e are all round.

Bhf= FED is required

and bhf= bhc+90°= bgc+90°fed= fea+90°= f'EA+90° so you need to prove F'e= bgc, i.e. e, f', g, c four points are circular and ae ac=ab =fa ga=f'a ga certified.

If you don't give a picture, it's tiring to watch. DP GM, EP GN written above, what's going on? Crossing the center of gravity g to make two straight lines intersect ac......? Anyway, it makes people see unclear and vague. So, I hope you will fill in the picture so that people can answer!

Something is wrong with the diagram above, doesn't the GN straight line pass through point D?

Provide a dumb way to spend an hour or so.

Establish a planar Cartesian coordinate system, set b(0,0) ,a(2b,2c) ,c(2a,0) and then you can calculate it slowly, anyway, you will definitely be able to prove it.

I wish you the best of luck from a purely geometric point of view.

<> connect PR and QR, extend BR to AC to D, pass E as the perpendicular line of BC, the perpendicular foot is N, and pass the perpendicular line of E as Ab, and the perpendicular foot is M

1) BPQ is an isosceles triangle, which is very easy and not written.

2) PBQR four-point contour, you need to write slowly.

First of all, you need to find the values of nq a1q and mp c1p.

Then find the value of er hr, there is an identity proportional relationship:

Where: <>

Available: <>

Synthesized above.

And because of that, en a1h, em c1h

Therefore, from the similar proportional relationship, we know that en rq, em rp, and then there are rq bc, rp ab, pbqr four points of the circle.

The question is so messy, I didn't understand it.

Where does the bisector of aa1 cc1 come from?

This question is relatively easy to use the analytical method, and the basic idea is as follows:

The equation for a circle is x 2 + y 2 = 1 (unit circle).

Let the coordinates of the q point be (m,n), then the tangent chord EF:MX+ny=1 makes the coordinates of point A (cos,sin), then the secant aq:y-n=[(sin -n) (cos -m)]*x-m)).

The M coordinates are obtained by the simultaneous tangent chord EF and the secant AQ.

The coordinates of the Q point are obtained by the simultaneous secant aq and the circle O.

Using the fixed-point formula, compare the coefficients of M and Aq of D. If it is equal, the conclusion is proven.