Eight, (full mark for this question 14)

23. As shown in Figure (L), convex quadrilateral ABCD, if point P satisfies ∠ APD = ∠ APB = α. And ∠ bpc = ∠ CPD = β, then point P is called the semi-equidistant point of quadrilateral ABCD.

(l) Draw a semi-equiangular point p in the square ABCD in Figure (3) and satisfy α ≠ β.

(2) Draw a semi-equiangular point p on the quadrilateral ABCD in Figure (4), and keep the drawing trace (without drawing method).

(3) If the quadrilateral ABCD has two semi-equidistant points P 1 and P2 (as shown in Figure (2)), it is proved that any point on the line segment P 1 P2 is also its semi-equidistant point.

Question 23

Jinan senior high school entrance examination in 2006.

27. (9 points for this question) As shown in figure 1, it is known that,,, and are connected at the point.

(1) length;

(2) Take the point as the center and the radius as the work, try to judge whether it is tangent to the point and explain the reason;

(3) As shown in Figure 2, if you click, you will drop your feet. Take the point as the center and the radius as the center; Take this point as the center and this point as the radius. If the sum is variable in size and keeps tangency in the process of change, the range of summation is made with points inside and points outside.

Nanchang city, Jiangxi province, graduated from junior high school in 2006 and took the entrance examination for technical secondary school.

25 background of the problem; At a study seminar, the extracurricular study group got the following two propositions:

(1) as shown in figure 1, in the regular triangle ABC, m and n are points on AC and AB respectively, and BM and CN intersect at point O. If ∠ bon = 60 and BM=CN:

② As shown in Figure 2, in a square ABCD, m and n are points. Bm on CD and AD respectively.

Intersect with CN at point o, if ∠ bon = 90, BM=CN.

Then, using similar ideas, the following propositions are put forward:

③ As shown in Figure 3, in the regular pentagonal ABCDE, m and n are points on CD and DE respectively, and BM and CN intersect at point O, if ∠ bon = 108, BM=CN.

Task requirements

(1) Please choose one of the three propositions ①, ② and ③ to prove it;

(Note: ① choose right 4 points, ② choose right 3 points and ③ choose right 5 points)

(2) Please continue to complete the following exploration;

① As shown in Figure 4, in the regular n(n≥3) polygon ABCDEF, M and N are points on CD and DE, respectively, and BM and CN intersect at point O. When ∠BON is equal to how many degrees, the conclusion BM=CN holds (without proof).

② As shown in Figure 5, in the regular pentagonal ABCDE, m and n are points on DE and AE respectively, BM and CN intersect at point O, ∠ bon = 108, is the conclusion that BM=CN still valid?

Yes, if so, please prove it. If not, please explain why.

(1) I choose

certificate

Nantong junior high school graduation entrance examination in 2006

(Question 28: 12) 28. As shown in the figure, in the plane rectangular coordinate system, O is the coordinate origin, B (5,0), M is the point on the OBCD of the isosceles trapezoid bottom, OD = BC = 2, ∠ DMC = ∠ DOB = 60.

Find the analytical formula of straight line CB;

Find the coordinates of m point;

∠DMC rotates α clockwise around point M (30 A+R

d=a+r

a-r