Review Unit 24. 1
◆ In-class detection
1, as shown in the figure, is the diameter, chord and vertical foot. Starting from these conditions, the conclusion _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _. Write only two conclusions, without auxiliary lines.
2. As shown in the figure, if the diameter is ⊙O and the point is two points on the circle, then _ _ _ _ _.
3. As shown in the figure, AD is the diameter ⊙O, AC is the chord, OB⊥AD, if OB=5 and ∠ CAD = 30, BC = _ _ _ _ _ _
4. As shown in the figure, AB=AC, ∠ APC = 60 is known.
(1) Prove that △ABC is an equilateral triangle. (2) If BC=4cm, find the radius ⊙ o. 。
◆ Typical case analysis
As shown in the figure, the chord length of the triangle-tangent circle O of △ABC is equal, ∠A=800, and the number of times to find ∠BOC.
C
B
A
O
Analysis: This problem is often easy to solve. Because the concepts of fillet and central angle are not well understood, it is easy to mistake ∠A for fillet and get ∠BOC=2∠A= 1600. In this problem, we should make full use of the condition that the chord length of △ABC three-sided truncated circle O is equal. We can get
Solution: the chord length of ABC's three-sided circle o is equal,
∴ The distance from the center O to the three sides is equal, and ∴0 is the center, that is, OB and OC share equally ∠ABC, ∠ACB.
∵∠a=800,∴∠abc+∠acb= 1000,∠obc+∠ocb=(∠ABC+∠ACB)= 500，
∴∠BOC= 1300.
◆ Homework after class
● Expand and upgrade.
1, as shown in the figure, is the two chords of a circle, and is the diameter of the circle, which is equally divided. The following conclusion is not necessarily correct ().
A, B, C, D,
B
D
C
A
2. As shown in the figure, if the length of the chord is cm and the distance from the center of the circle is 4cm, the radius of the chord is ().
A, 3cm b, 4cm c, 5cm d, 6cm
3. As shown in the figure, if AB is ⊙O in diameter, C, D and E are all on ⊙O, and ∠ bed = 30, then ∠ACD is ().
A.60 B.50 C.40 D.30
4. As shown in the figure, in ⊙O, the chord AB and CD intersect at point E, ∠ BDC = 45, ∠ Bed = 95, then the degree of ∠C is _ _ _ _ _.
A
B
C
D
E
O
5. As shown in the figure, in △ABC, ∠ ACB = 90, D is the midpoint of AB and the ⊙O intersection with DC as the diameter.
The edges of △ABC are at G, F and E points.
It is proved that (1)F is the midpoint of BC; (2)A =∠GEF。
A
B
C
D
E
F
G
O
6. As shown in the figure, ⊙O and ⊙O intersect at point A and point B, the moving point P is on ⊙O, and outside ⊙, the straight lines PA and PB intersect at point C and point D ⊙O respectively. Q: Does the length of the chord CD of ⊙O change with the movement of point P? If there is any change, please determine the position of the longest and shortest CD P. If there is no change, please give proof.
● Experience the senior high school entrance examination
1, (Shaoxing, 2009) As shown in the figure, in the plane rectangular coordinate system, ⊙P and x axes are tangent to the origin o, and the straight line parallel to y axis intersects ⊙P at m and n points. If the coordinate of m point is (2,-1), the coordinate of n point is ().
a 、( 2，-4) B 、( 2，-4.5) C 、( 2，-5) D 、( 2，-5.5)
2. (Putian, 2009) (1) Draw according to the following steps and mark the corresponding letters: (Draw directly in the figure 1)
① Draw a semicircle with the diameter of the known line segment (Figure1);
② Take a point different from the point on the semicircle and connect it;
(3) Draw a semicircle at this point.
(2) Drawing with a ruler: (traces of drawing are reserved, and writing methods and proofs are not required)
Known: (Figure 2).
Find the bisector of.
Reference answer:
◆ In-class detection
1, AC=BC, etc.
2、40 .
3、5.
4.( 1) Proof: ∫∠ABC =∠APC = 60,
Also, ∴∠ ACB = ∠ ABC = 60, ∴△ABC is an equilateral triangle.
(2) solution: connect OC, the intersection o is OD⊥BC, and the vertical foot is D,
In Rt△ODC, DC=2 and ∠ OCD = 30,
Let OD=, then OC=2, ∴oc=.∴.
◆ Homework after class
● Expand and upgrade.
1、a。
2、c。
3、a。
4、40 .
5. It is proved that (1) connects DF, acb = 90, D is the midpoint of AB, ∴ BD = DC = AB,
∵DC is the diameter ⊙O, ∴ DF ⊥ BC. ∴ BF = FC, that is, F is the midpoint of BC.
(2)d and F are the midpoint of AB and BC respectively, ∴DF∥AC, ∠ A = ∠A=∠BDF,
∵∠BDF=∠GEF,∴∠A=∠GEF.
6. Solution: When point P moves, the length of CD remains the same. The reason for this is the following：
Connect ads. ∵A and B are the intersection of ⊙O and ⊙O, and ∴ chord AB has nothing to do with the position of point P 。
ADP is in ⊙.
∵∠P is in∵∣∣∣
∫∠CAD =∠ADP+∠p, ∴∠CAD is a fixed value, and in ∫∝8.
The length of CD has nothing to do with the position of point p.
● Experience the senior high school entrance examination
1, B. Using the vertical diameter theorem.
2. Solution: (1) omitted.
(2) Take this point as the center and an appropriate length as the radius, and make an arc intersection between two points.
Take this point as the center and the radius greater than the length as the arc, and the two arcs intersect at this point.
Emit rays.