Theme explanation

In middle school mathematics, the position relationship of figures mainly includes the relationship between points, lines, triangles, rectangles, squares and circles. The senior high school entrance examination will be included in the problems of function, coordinate system and geometry, but mainly through the relationship between circle and other figures, among which the most important ones are various problems of circle and triangle. Because this kind of problems are basically the second way to solve the upper-level problems, the difficulty is generally above average after calculating the angle of the line segment. Therefore, how to take the score of this question into your arms has become a problem that every candidate and parent has to pay attention to. From the topic itself, it is generally a very standard two-question formula. The first question proves the tangent, examines the tangent judgment theorem, tangent property theorem and inference. The second question usually gives the length of a line segment and the trigonometric function value of an angle, finds the length of other line segments, and comprehensively examines the knowledge points of circles and triangles. As for other graphic position relationships, we will introduce them in later topics.

Illustration

Example 1 known: As shown in the figure, AB is the diameter ⊙O, and ⊙O passes through the midpoint D of AC, and De ⊥BC is at the point E. (1). Prove that DE is the tangent of ⊙O; (2) if DE =2, tan C =

1

Find the diameter of ⊙ o.2

A

In recent years, the thinking analysis shows that this kind of problems especially like to take the midpoint problem into account. Candidates must be very sensitive to the parallelism caused by the midpoint and the midline, especially don't forget that the center of the circle is also the midpoint of the diameter. For this problem, the natural connection OD, that is, the midline in △ABC, is parallel to BC. So OD ⊥DE can be proved by vertical transfer relation. As for the second question, the focus is on the knowledge point with a diameter of 90. By using perpendicular bisector relation, it is concluded that △ABC is an isosceles triangle, thus the solution of AB is transformed into BD, and the problem of circle is transformed into the problem of solving a right triangle, so it is easy to get the solution. analyse

(1) proves the connection OD. D is the midpoint of communication, and O is the midpoint of AB.

∴ OD is the center line of △ABC. ∴od∨BC。 * ∴, BC 1990. ∴∠ Ode = ∠ DEC = 90 degrees. ∴

∴ De is the tangent of⊙ O.

(2) Solution: Connect the database. ∵ ab is the diameter⊙ O,

∴∠ ADB = 90. ∴ DB ⊥ AC。 ∴∠ CDB = 90。 ∫d is the midpoint of communication, ∴ AB = communication.

In Rt△DEC, DE = 2 and tanC= are obtained by pythagorean theorem:

DC=

∴ec==4. 1de (remember the meaning of trigonometric function) 2tan C

In Rt△DCB,

BD=DC？ tan C = BC=5。 ∴AB=BC=5.

∴⊙O has a diameter of 5.

Example 2 is known: As shown in the figure, O is? The circumscribed circle of ABC, where BC is the diameter of O, and is taken as ray BF, so that BA bisects ∠CBF, and point A is taken as AD ⊥BF of point D. 。

(1) verification: DA is the tangent of o; (2) if BD = 1, tan ∠BAD =

F 1

, find the radius of o.2.

C

The problem of train of thought analysis is a typical problem of proving tangent with angle. In the title, except for the vertical relation, only a BA bisector ∠CBF is given. Seeing this situation, we need to realize that we should prove parallelism from the angle. To prove parallelism by angle is nothing more than that the isosceles angles of internal dislocations are equal and the internal angles on the same side are complementary. In this problem, even after OA, it is found that ∠ABD=∠ABC and an isosceles triangle are formed, so that ∠ABO=∠BAO naturally wants to convey the relationship between these angles, and thus proves it. The second problem is to use the angle transfer to put the known angle ∠BAD into △ABC through the equal relationship, so as to achieve the purpose of calculating the diameter or radius. Analysis proves that: connecting AO.

ao = bo，∴ ∠2=∠3。 ∫ba shares ∠CBF, ∴∠ 1 = ∠ 2. ∴∠ 3 = ∠.

∴db∑ao。 (Score points, don't forget to prove the parallelism of equal internal angles) ∵ AD ⊥DB

∴ ∠BDA =90？ . ∴∠ Road =90? . ∵ AO is the radius⊙ o,

∴ DA is the tangent of ∴ O. (2) ∴ ad ⊥ db, BD = 1, tan∠ bud =∴ AD =2.

Obtaining ABⅶ from pythagorean theorem.

sin ∠4=

(Extending known conditions through transformation of trigonometric functions) f

C

1, 2

∵ BC is the diameter⊙ O,

∴ ∠BAC =90？ . ∴ ∠C +∠2=90？ And ∵ ∠4+∠ 1=90? , ∠2=∠ 1,

∴∠ 4 =∠ C. (In this step, BD/AB=AB/AC=sin∠BAD can also be directly derived from triangle similarity).

In Rt △ABC, the radius of BC =∴ O is

5.2

AB AB

==5.sin C sin ∠4

Example 3 is known: As shown in the figure, point D is a point on the extension line with diameter CA≥O, and point B..

On ⊙O, and OA =AB =AD. (1) proves that BD is the tangent of ⊙O;

(2) If point E is a point on the lower arc BC, AE intersects BC.

At point f, BE =

8. tan ∠BFA = Find the radius length of ⊙ O. 。

C

The train of thought analysis includes OA=AB=OD. Smart students can immediately see that BA is actually the center line on the hypotenuse OD of a triangle. Then, according to the inverse theorem of the theorem that the midline of the hypotenuse of a right triangle is equal to half of the hypotenuse, it can be deduced immediately that ∠ OBD = 90, so that the tangent problem is solved. In fact, if you can't see it, then connecting OB can be done by angle transmission as in Example 2. The second question of this question is a little more difficult, and some properties of the angle of circle are also examined. Prove the similarity of triangles with equal circle angle, and connect the unknown conditions with the known conditions in proportion. In recent years, the scope of the senior high school entrance examination has been compressed, and other extracurricular contents such as the circular power theorem are basically not needed, so similar triangles is more used for proportional calculation. I hope you can master it carefully. Analytic (1) proof: join OB.

oa = ab，OA =OB，∴OA =AB =OB。

∴? ABO is an equilateral triangle.

∴∠BAO =∠ 1=60？ . ab = ad，∴∠D =∠2=30？ .

∴∠ 1+∠2=90? .

∴DB ⊥BO. (It's just a bit troublesome to solve this problem without the inverse theorem of hypotenuse midline) And ∵b is on ⊙O, and ∴DB is the tangent of ⊙ o. 。

(2) Solution: ∵CA is the diameter ⊙O, ∴∠ABC =90? .

At Rt △

In ABF, Tan ∠BFA =

∴ set AB,

AB =BF C

Then BF =2x,

∴af = 3 times. ∴

Boron difluoride

=. (the idea of setting elements is very important) AF 3

∠∠c =∠e，∠3=∠4，∴？ BFE ∽？ AFC. ∴

Do BF 2

== .

AC AF 3

∵BE =8，∴AC = 12。

∴AO =6。 ? 5 points

Example 4 as shown in the figure, in the isosceles triangle ABC, AC =BC =6, AB = 8. Take BC as the diameter, let O and AB intersect at point D, AC intersect at point G, DF ⊥AC, the vertical foot is F, and the intersection point CB is the extension line of point E. (1) Prove that the straight line EF is the tangent of O; (2) Find the value of sin ∠E 。

The difference between this problem and the previous one is that it is calculated and proved by the specific length of the line segment. If you want to prove that EF is tangent, you need to prove that OD is perpendicular to EF, but this question does not give the relationship between OD and other line angles, so you need to make an additional auxiliary line to connect CD, and use a circumferential angle with a diameter of 90. △ABC is an isosceles triangle with AC and CB as the waist, so D is the midpoint. Successfully transformed into the previous midpoint problem, and then solved. The second question uses the result of the first question to translate the known angle and construct the algebraic relationship between similar RT triangles with the help of Pythagorean theorem.

The formal solution of Equation A also examines the examinee's efforts to solve the triangle.

Analysis (1) proves: As shown in the figure, if a CD is linked, then ∠BDC =90? .

F

∴CD ⊥AB。 D

ac = bc，∴ AD = BD。 ∴ D is the midpoint of AB ∵ O is the midpoint of BC, ∴do∨AC. EF ⊥ AC is at EF ⊥ Do.

∴EF is the tangent of o

(2) Connect BG, ∫BC is the diameter, ∴∠BGC =90? = ∠ cfe。 (Diameter and circumferential angle are all 90) ∴ BG ∑ ef.

FC CG

∴sin ∠E =。 =

EC BC

Let CG =x, then ag = 6-X.

In Rt △BGA, BG 2 = BC 2-CG 2. In Rt △BGC, BG 2 = AB 2-AG 2. (This step is very important. The adjacent sides of two adjacent RT △ can be used to construct the equation, and in fact, the hypotenuse of a right triangle can be directly used to obtain a high proportion. )

222

∴62 x 2 = 82-(6 x)。 The solution is x = .. which means CG = ..

33

At Rt △BGC.

2CG 1

==.∴ sin ∠E =

69 BC

Example 5: As shown in the figure, in the parallelogram ABCD, the circle with A as the center and AB as the radius intersects with AD in F, BC in G and BA in E. 。

(1) If ED is tangent to ⊙A, try to judge the positional relationship between GD and ⊙A and prove your conclusion;

(2) If the condition of (1) remains unchanged, if GC = CD = 5, find the length of AD.

Food and drug administration

University of Cambridge, UK

Although this topic is about the positional relationship between a circle and a parallelogram, it still examines the ability to solve all conditions in the most basic triangle. It is not difficult to judge that DG is tangent to the circle, but how to prove it. In fact, in addition to this topic, Mentougou, Shijingshan and Xuanwu have all investigated the proof that two tangents are drawn from a point outside the circle. The most important thing of this kind of problem is to prove the similarity of triangles by using the equal radii of circles and the equal central angles of two circles. The second problem is not difficult, and the key point is how to judge the special angle in the right triangle by using the double angle relationship, so as to solve it. analyse

(1) Conclusion: The tangency of GD and O proves that the points G and E connecting Ag: are on a circle, and the quadrilateral ABCD ∴ag = AE∶: is a parallelogram, and ∴ad∨BC.

∠2=∠3 ∴∠B =∠ 1，ab = ag

∴∠B =∠3

∴∠ 1=∠2 (If you do more, you will find that this kind of question is basically to find this diagonal line, so candidates should be good at grasping the known conditions and leading to it)? AED and? AGD

E

E

? AE =AG？

? ∠ 1=∠2 ? AD =AD？

A

12G

F 5C

D

∴? AED?？ Agd ∴∠ aed = ∠ agd ∠ ed and ∴∠aed = 90° tangent? ∴∠AGD =90？ ∴AG ⊥DG

∴GD is tangent to a

(2)∫GC = CD = 5, quadrilateral ABCD is parallelogram ∴AB =DC, ∠4=∠5, ab = ag = 5∫ad∨BC ∴∠4=∠6.

B

1

∴∠5=∠6=∠B

2

∴∠ 2 = 2∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠873 In fact, putting the multiplication relation in the RT triangle produces special angles of 30 and 60) ∴∠6=30?

∴AD = 10。

To sum up, we can get a general idea to solve this kind of problem. It is not necessary to prove tangency, and the center of the circle is connected with the tangent point as an auxiliary line. Next, we should consider how to prove that the radius is the distance from the center of the circle to the tangent, that is, "connecting radius proves verticality". In recent years, the senior high school entrance examination basically only requires this idea of proving tangency, but there are actually three ways to prove tangency. In case of encounter, I still hope candidates can understand.

The first is to connect the radius first and then prove the verticality in the textbook. The premise of this is that the conditions given in the question have implied the radius. For example, a circle circumscribes a triangle, or the intersection of a circle and a line segment. As long as we have a good grasp of the nature of various circles.

The second is that if the topic does not give the intersection state, it can't be connected rashly. You can make a vertical line first, and then prove that the vertical line is equal to the radius, which is the so-called "proving the vertical before proving the radius". For example, look at such a problem, as shown in figure △ABC, AB=AC, point O is the midpoint of BC, and point D is tangent to AB, which proves that it is also tangent to AC.

Whether the circle 0 and AC have something in common in this problem is unknown, so we can only make the vertical line of AC through O, and then prove that this distance is exactly the radius of the circle. If candidates take it for granted that there is an intersection, and then directly connect the intersection of AC and circle to prove it, they will go astray.

The third one is more difficult. On the one hand, the relationship between radius and vertical line is not given in the topic, which belongs to the question type that both radius and vertical line should be proved. For example, look at the following question: As shown in the figure, the middle AB=AC, =, O, D divides BC into three parts, and draws with OB as the center to prove that it is tangent to AC.

This question does not specify that it must pass through point A, so it is very troublesome to prove that A is the tangent point and that the vertical foot from O to AC coincides with A. But to put it another way, if you connect AO and prove that AO=OB, AO ⊥AC, it is very strict.

(Hint: If you make a vertical line, then the vertical foot is also the midpoint. We can prove that AO and BO are equal through the quantitative relationship represented by AB, and in △AOC, we can prove the right angle by the inverse theorem that the length of the hypotenuse of a right triangle is half that of the hypotenuse.

As for the calculation of the second question in this kind of question, it is relatively simple. Grasping the line segment and angle relationship formed by the central angle, the central angle and the possible tangent angle, and putting the conditions in the same RT △, the solution can be very convenient.

In short, this kind of topic is not too difficult, and we need to do it quickly and with high accuracy, so as to leave room for future generations.

Hand training

1. As shown in the figure, it is known that AB is the chord of ⊙O, C is the upper point of ⊙O, ∠C =∠BAD, BD ⊥AB proves in B. (1) that AD is the tangent of ⊙O;

(2) If the radius ⊙O is 3 and AB =4, find the length of AD.

Proof: As shown in the figure, if AO is connected, the intersection ⊙O is extended to point E, and BE is connected, then ∠ Abe = 90. ∴∠ EAB+∠ E = 90。

∠∠E =∠C，∠C =∠BAD，

∴∠ ∴ ∠EAB +∠BAD =90。 ∴ AD is the tangent of∵ o.

(2) Solution: (1) shows that ∠ Abe = 90.

AE = 2AO = 6，AB =4，

∴ BE =AE 2-AB 2=25。 ∠∠e =∠c =∠bad，BD ⊥AB，∴ cos ∠BAD =cos ∠E。

AB BE ∴ =。

account executive

that is

42=. AD 6

12∴ Advertising =

five

2. As shown in the figure, AB is the chord of ⊙O, the intersection o is the parallel line of AB, the intersection ⊙O is at point C, and point D on the straight line OC satisfies ∠D =∠ACB.

(1) Judge the positional relationship between straight line BD and ⊙O, and prove your conclusion; (2) If the radius of ⊙O is equal to 4, tan ∠ACB = solution: (1) Proof that the straight line BD is tangent to ⊙ o: As shown in Figure 3, the connecting line OB. -

∠∠OCB =∠CBD+∠d，∠ 1=∠D，∴ ∠2=∠CBD。 ∫ab∥oc，∴ ∠2=∠A . ∴ ∠A =∠CBD。 OB = OC，

∴ ∠BOC +2∠3= 180？ ，∫∠BOC = 2∠A，

∴ ∠A +∠3=90？ . ∴ ∠CBD +∠3=90？ . ∴ ∠OBD =90。

The straight line BD is tangent to O.

four

Find the length of the CD

D

(2) solution: ∫∠d =∠ACB, tan ∠ACB =∴ tan D =

4.3

4, 3

In Rt △OBD, ∠ OBD = 90, OB = 4, tan D =∴ sin D =

40b，OD ==5。 5sin D

four

, 3

∴ CD =OD -OC = 1。

3. As shown in the figure, in △ABC, AB=AC, AE is the angular bisector, BM bisector ∠ABC and AE intersect at point M, ⊙ O passing through points B and M intersects with BC at point G, AB intersects with point F, and F and FB are just the diameter of ⊙ O. (1) Verification.

1

When, find the radius ⊙ o.3

1) Proof: If OM is linked, then OM = ob. ∴∠ 1 = ∠ 2.

∵BM shares ∠ ABC ∴∠ 1 = ∠ 3.∴∠ 2 = ∠ 3.∴om∨BC。

∴∠AMO =∠AEB。

In △ABC, AB =AC, AE is the angular bisector, ∴ AE ⊥ BC. ∴∠ AEB = 90。 ∴∠ Amos = 90. ∴ Om ⊥ Love. Ae and

(2) solution: In △ABC, AB =AC, AE.

1

∴BE =BC，∠ABC =∠C。

2

1

cos C =，BC = 4，

three

1

cos ∠ABC =。 ∴BE = 1，

three

At △ABE, ∠ AEB = 90,

exist

=6.∴AB =

cos ∠ABC

Let the radius of ⊙O be r, then ao = 6-r .∵om∨BC,

∴△aom∽△ Abe. Omo

Become a member of = = .26

(2) When BC=4, cosC=

B

The solution is r =

3.2

3.2

∴⊙O has a radius of

4. As shown in the figure, in the isosceles △ABC, AC=BC, ⊙O is the circumscribed circle of △ABC,

As far as the last point is concerned, ⊥AD was in 400 BC.

Verification: AE= BD +DE ..

Proof: As shown in Figure 3, intercept AF=BD on AE and connect CF and CD.

In △ACF and △BCD,

? AC =BC，∠CAF =∠CBD，？ AF =BD，？

∴△ACF?△BCD。 ∴ CF=CD。

CE ⊥AD, ∴ EF=DE in e. ∴ AE =AF +EF =BD +DE。

5. As shown in the figure, it is known that ⊙O is the circumscribed circle of △ABC, AB is the diameter of ⊙O, and D is AB.

The extension line from AE ⊥CD to DC is in E, CF ⊥AB is in F, and CE = CF 。

(1) Verification: DE is the tangent of ⊙O;

(2) If

AB = 6, BD = 3, find the length of AE and BC. It is proved that (1) connected oc,

AE ⊥CD, CF ⊥AB, and CE = CF, ∴∠ 1 = ∠ 2. OA = OC，∴∠ 2 = ∠ 3。 ∴∠ 6544; .

Christian era

(2) solution: AB =6, ∴OB =OC =

1

AB =3。 2

At Rt? In OCD, OC =3, OD =OB +BD =6, in Rt? In ADE, ad = ab+BD = 9, AE =

19AD =22

Are you online? In OBC, ∠COD=600, OB =OC.

∴∠D =300。 ∠COD =600。

∴BC =OB =3。

Outdoor training

1.(20 1 1 Nanjing, Jiangsu) As shown in the figure, in Rt △ABC, ∠ ACB = 90, AC =6㎝, BC =8㎝, and P is BC.

Midpoint. The moving point Q starts from point P and moves at a speed of 2㎝/s along the direction of ray PC to make a circle with P as the center and PQ length as the radius. The time to move point q is t s.

(1) When t = 1.2, judge the positional relationship between straight line AB and ⊙P, and explain the reasons; ⑵ O is known as the circumscribed circle of △ABC. If υp is tangent to υo, find the value of t 。

(Question 26)

The positional relationship between a straight line and a circle refers to intersection, tangency and separation, and the judgment is based on the distance between the straight line and the circle, so only the distance from the center p to the straight line AB when t= 1 is required. 2. There are many positional relationships between two circles. Because point P is in a circle, it is enough to contain, intersect and cut. The basis of judgment is the distance between the centers of two circles. At the same time, we should pay attention to all kinds of possibilities and make the answer comprehensive. The answer (1) The straight line AB is tangent to ⊙ p. 。

As shown in the figure, the intersection point P is PD ⊥AB and the vertical foot D. At Rt △A BC, ∠ ACB = 90, ac = 6cm and BC =8cm.

∴ AB = 10 cm. ∵ P is the midpoint of BC, ∴ Pb = 4 cm.

∠∠P DB =∠ACB = 90，∠PBD =∠ABC。 ∴△PBD ∽△ABC。

∴PD PB PD 4==, that is ∴ PD = 2.4 (cm). ACAB6 10。

When t = 1.2, PQ =2t =2.4(cm).

∴PD =PQ, that is, the distance from the center of the circle P to the straight line AB is equal to the radius \u p 。

The straight line AB is tangent to p.

5] ⑸ACB = 90°, ∴AB is the diameter of the circumscribed circle of △ABC. ∴ OB =

The line op .∫p is the midpoint of BC, ∴ op =1ab = 5cm.21AC = 3cm.2.

∵ Point P is within ⊙O, ∴⊙P and ⊙O can only be inscribed.

∴5-2t =3 or 2t -5=3, ∴t = 1 or 4.

When ∴⊙P is tangent to O, the value of t is 1 or 4.

2. As shown in the figure, AC is the diameter of ⊙O, PA ⊥ AC and BC are chords of ⊙ O, and the straight line PB intersects with the straight line AC at point D, = = .dpdo3

(1) Verification: the straight line PB is the tangent of ⊙O; P (2) Find the value of cos ∠BCA. b

Office of the secretary of defense

The problem-solving idea (1) proves that the tangent and radius must be connected, and proves that it is vertical. Connect OB and OP, prove? BOP？ AOP is enough;

In the second sub-question, the problem should be transformed into the cosine of POA by parallel lines, and the cosine of POA should be found in Rt? Prison prison association

Let PA =a, according to the known conditions, use an algebraic expression containing a to represent the side lengths OA and OP, and then use a triangle.

Find by function.

The answer (1) proves: connect OB, op ............................. (1) ∫ = = and ∠D =∠D DP DO 3.

∴△BDC ∽△PDO

P ∴∠DBC =∠DPO

∴bc∑op

∴∠BCO =∠POA

∠CBO =∠ balance of payments D A C O \o b = OC

∴∠ocb =∠ Congressional Budget Office

∴∠BOP =∠POA

OB = OA OP = OP again

∴△bop?△AOP

∴∠PBO =∠P AO

Once again: \p a ⊥ac

∴∠PBO =90

∴ straight PB and to⊙∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴8

(2) It is known from (1) that ∠BCO =∠POA.

Let PB =a and BD =2a.

PA = PB = a

1 1

∴AD =

∫BC∑op ∴dc = 2。

∴DC =CA =？ =

2

∴OA

∴OP = ∴cos

BCA = cos

(20 1 1 Yongzhou, Hunan) As shown in the figure, AB is the diameter of a semicircle O, and point C is a point above ⊙O (A and B do not coincide), connecting AC and BC, and the intersection point O is OD∑AC intersects BC at point D, taking a point E on the extension line of OD and connecting EB to make it ∞.

(1) Verification: BE is the tangent of ⊙O;

⑵ If OA= 10 and BC= 16, find the length of BE.

B

Thinking of solving problems: (1) To prove that BE is the tangent of ⊙O, it is necessary to prove that ∠ OBE = 90. According to the fact that AB is the diameter of a semicircle O, we can get ∠ ACB = 90, and deduce ∠ CAB+∠ ABC = 90, and then get it by parallel. (2) To find the length of BE, firstly calculate the length of AB according to Pythagorean theorem, and then calculate the length of BE by using acute trigonometric function or similar methods.

The answer proves that (1) ∵ AB is the diameter of a semicircle with ∴∠ACB = 90°.

∫od∑AC ∴∠odb=∠acb=90 ∴∠bod+∠abc=90

∠∠oeb =∠ABC∴∠BOD+∠oeb = 90∴∠OBE = 90。

∵AB is the diameter of the semicircle O, ∴BE is the tangent of∵ O.

(2) at Rt？ In ABC, AB=2OA=20, BC= 16, ∴ AC = AB 2-BC 2 = 202-162 =12 ∴ Tana =

∴BE =

(Topic 25) BC164be4 = ∴ tan ∠ boe = = AC123ob3441ob =? 10= 13.

333

12