(1) Please find the similar triangles in the diagram and prove them; F
(2) If BC=2, find the area. o e d。
Solution: (1) Conclusion:. Evidence: in a word, BC
Figure 5- 15
(2) The diameter BF as shown in the figure, connecting CF, is the area of an isosceles right triangle.
2. (Mianyang, 2005 ★★) As shown in Figure 5- 16, it is known that BC is the diameter, while the vertical foot is D, point A is the midpoint, BF intersects with AD at point E, and. a F。
(1) verification: AE=BE E
(2) Find the length of DE; British overseas company
(3) Find the length. Director of Buildings
Solution: (1) linkage AF, where a is the midpoint of BF,
And ... H Figure 5- 16
BC is the diameter, and.
(2) Let AD=6 exist,
The length of DE is 2.
(3) From (1), (2) BE=AE=6-2=4, in,.
3.(2004. Shanxi ★★★★) As shown in Figure 5- 17, the straight line CD which is located above and passes through Point A and intersects with Point A and Point B, the straight line EF which intersects with Point C and Point D respectively and the chord D which intersects with Point E and Point F intersect at Point P..
Verification: (1) ce//df; (2) .
It is proved that (1) quadrilateral ABEC is an inscribed quadrilateral,
The quadrilateral ABFD is.
Inscribed quadrilateral,.
(2) Link, and then ... Figure 5- 17.
Here we go again.
Namely.
4.(2005. Hai 'an ★★★★★) As shown in Figure 5- 18, it is known that O is a point on the diagonal of the square ABCD, with O as the center, and the length of OA is the radius tangent to BC at point M, which intersects AB and AD at points E and F respectively.
(1) Verification: CD and tangency;
(2) If the side length of the square ABCD is 1, find the radius of;
(3) For the tangents of points M, E, A, F and CD, it is
The five sides of a Pentagon with a vertex, considering the equal relationship, can
What is the conclusion? Please prove it.
Proof: (1) As shown in Figure 5- 18, connect OM to make it in N. 。
Tangent to BC .. quadrilateral ABCD is a square,
AC bisection, CD and tangency. Figure 5- 18
(2) The quadrilateral ABCD is a square with AD=CD= 1,
NC=ON=OA，。
, .
(3)ME=FN，AE=AF。 As shown in figure 5- 18, divide evenly with AC.
. , tangent to,,,
and
5. (Wenzhou, 2005) As shown in Figure 5- 19, quadrilateral ABCD is known.
A is the midpoint of the best arc. At point A,
And the extension lines of CB intersect at point f and point e respectively,
EM cuts at point m.
(1) Verification:; Figure 5- 19
(2) Verification:
(3) If AB = 2 and EM = 3, find the value of.
It is proved that (1) quadrilateral ABCD is inscribed in,.
.
(2) A is the midpoint of point H (as shown in figure J 1 1-2).
,
(3) A is the midpoint, AB=2, AC=AB=2. EM is the tangent, EM=3,.
. ② ①+②
6 As shown in the figure, points A, B, C and D are above, and point E is on the extension line of DC, ∠ BOD = 120, then
The degree of ∠BCE is _ _ _ _ _.
& lt answer > 60
< Analysis > Connecting AB, AD found that quadrilateral ABCD is an inscribed quadrilateral, while ∠BCE is a quadrilateral ABCD.
The external angle of, so ∠BCE=∠A, and then by the theorem of circumferential angle ∠ BCE = ∠ A = ∠ BOD = 60.
< Comment > This question examines the inscribed quadrilateral of the auxiliary line as a circle, and then applies knowledge point 2(2)(7) to solve it.
Fig. 7 shows the broken part of the circular grinding wheel. Try to find its center.
< Analysis > According to the vertical diameter theorem and its inference, it is relatively simple.
< Answer > Solution: Practice:
(1) Take one point for any post;
(2) Connect AC and BC, and make the sum of the vertical lines of AC and BC respectively, and the sum intersects at point O, then point O is the center of the circle.
The most important property of a circle is symmetry, and the vertical diameter theorem can be regarded as a conclusion drawn from symmetry.
8(2005? Suqian city) is known: as shown in the figure, in △ABC, AC = BC. The intersection AB with the diameter of BC is at point D, and the intersection D is the DE AC of point E. The extension line of the intersection BC is verified at point F: (1) ad = BD; (2)DF is tangent.
& lt analysis > (1) In order to prove that AD=BD, we need to consider the positions of two line segments. AD and BD are on the bottom of isosceles △ABC, so we should consider starting from the nature of isosceles triangle, using the nature of the perpendicular line, or using the congruence of triangle.
(2) To prove the tangent of DF, first follow OD, and try to prove that ODF is equal to a known right angle, or ODF is.
& lt answer > (1) Method 1: Link the CD,
BC is the diameter.
AC=BC。 AD=BD。
Method 2: connect the CD,
BC is the diameter.
AC=BC，CD=CD。 △ACD △BCD。
AD=BD
(2) method 1: connecting od,
AD=BD，OB=OC。 OD//AC。
DE AC, DF is tangent.
Method 2: Connect od.
OB=OD。 .
. .
DF is tangent.
< Comment > This topic examines the knowledge of triangles, the application of knowledge of tangents and circumferential angles in circles, and the simple reasoning and proof of geometry.
9 In the rectangular coordinate system, the center of ⊙O is in a point with a radius of 3, the coordinate of the center of ⊙A is (-,1), and the radius is 1, so the positional relationship between ⊙O and ⊙A is ().
A, circumscribed b, circumscribed c, inscribed d, intersecting
& lt answer > c
< Analysis > Through calculation, it can be concluded that the distance between the two centers of the circle OA=2 and the radii of the two circles are 1, 3 respectively, so the positional relationship between the two circles is determined to be inscribed.
& ltExpand > What is the answer if the position of the center O in this question is changed to (1, 2)?
10 (Haidian, Beijing, 2002) As shown in the figure, AB is the diameter ⊙O, AE bisects ⊙O at point E, and the intersection point E is a straight line perpendicular to AF. The AF extension line is at point d, and the AB extension line is at point C.
(1) Verification: CD and ⊙O are tangent to point E;
(2) If, find the tangent of the sum of diameters ⊙ o.
& lt answer > solution: (1) prove 1: connect OE.
Divide equally,
,
( 1)
,
Proof (2 points)
It's a point on ⊙O,
Tangent to O at point E (3 points)
Proof 2: Connect BF and OE at G point.
Divide equally,
AB is the diameter.
( 1)
The following is the same as the first method.
Evidence 3: Connecting BE and OE
Divide equally,
AB is the diameter,
( 1)
,
(2 points)
E is the point on ⊙O,
CD is tangent to O at point E (3 points).
(2) Solution 1:
Intersection d makes DG//AC cross AE extension line at G point, connecting BE and OE.
Tangent to ⊙O, tangent to point e,
,
,
(5 points)
OE//AD reference (1)
That is all right
(6 points)
Pack up, take it
Solution (negative),, (7 points)
⊙ The diameter of O is (8 points).
From the cutting line theorem
Yes, (9 points)
Solution 2:
Join BE, OE, EF
certificate
namely
Tangent to ⊙O, tangent to point e,
This is another corner,
& lt2 & gt
Through < 1 >,<2>, we have to
,
(5 points)
The following is the same solution.
1 1 As shown in the figure, the pentagon ABCDE is a regular pentagon, and the curve EFGHIJ… is called "the involute of the regular pentagon ABCDE", where the centers of,,, and … are a, b, c, d, E… cycles in turn, and they are connected in turn. If AB= 1, then.
& lt answer > six
12 As shown in the figure, the side length of the square ABCD is 10, and the point P is a point in the square. Connect PB, PB=6, and connect PA to get △PAB. Rotate △PAB clockwise around point B by 90 to △ P ′ CB, and find out the sweep range of PA side when rotating from △PAB to △ P ′ CB.
& lt Answer > Solution: Judging from the meaning of the question, = 90.
and
A: A little.
Exercise 8 There is mathematics everywhere in our life. The picture below shows the toilet paper roll commonly used in our life. Do you know how thick each layer of toilet paper is? From the packaging of toilet paper, we got the following information: "There are 300 cells in two layers, each cell11.4cm×11cm (length× width)". We measured the inner and outer diameters of the whole roll of toilet paper with a ruler, which were 2.3 cm and 5.8 cm respectively, and the thickness of each layer of toilet paper was about _ _ _ _.
& lt answer > 0.0 13