(1) With chords, the frequently cited auxiliary lines are: the radius of the end of the chord; The diameter perpendicular to the chord (or the distance from the chord center). Function: Form a right triangle or use the vertical diameter theorem. Memory formula: circle is not difficult to prove, radius and diameter are often connected; If there is a chord, if you want the chord center distance, it will split the chord vertically; Example 1: as shown in figure 1, AB is the chord of ⊙O, p is the point ABove ab, AB= 10cm, PA=4cm, OP=5cm. Find the radius of ⊙ o [canonical solution] Do OC⊥BA in C, followed by OA. Then in Rt△AOC and Rt△POC, AO2-AC2=OP2-CP2 is AO2-52=52-(5-4)2. ∴ AO=7. That is, the radius of∵ o is 7cm. Example 2: AB = CD is known, and M and N are the midpoint of AB and CD respectively. Verification: ∠AMN=∠CNM [canonical solution] M and N are OM⊥AB and ON⊥CD, respectively, and the vertical foot is M, n∶ab = cd,∴om=on, ∴∠omn =∞. ∵OM⊥AB、ON⊥CD∴∠OMA=∠ONC=90 ∴∠AMN=∠CNM。 (2) With the diameter, the frequently cited auxiliary line is: the circumferential angle corresponding to the diameter. Function as shown in the figure: get a right angle or a right triangle. Memory formula: Make a right angle when the diameter is satisfied Example 3: (Senior High School Entrance Examination in 2007) ①AD is the diameter of circle O, BC tangent circle O intersects circle O at D, AB, AC at E and F. Verification: AE ab = af AC. Revelation: AD is the diameter and the circumferential angle of the structure diameter. [canonical solution] the connecting line DE, df∶AD is the diameter of the circle o of df⊥ac. ∴de⊥ab ∫BC is tangent to the circle o of point d, and ad is the diameter of the circle o, ∴AD⊥BC.∴ According to the projective theorem, there are Ad2 = AE AB and AD2. ∴AE AB=AF AC. Example 4: It is known that ⊙O 1 and ⊙O2 intersect at point A and point B, and O2 is on ⊙O 1. AD is the diameter of ⊙O2, connecting DB and extending the intersection of ⊙O 1 to C, which proves that: CO2⊥AD. Revelation: AD is the diameter and the circumferential angle of the structure diameter. [canonical solution] the connection ab: ad is: O2 diameter ∴∠ ab∶ad is a right angle ∴∠ABC is a right angle ∠ABC and ∠A02C are circumferential angles on the same arc ∴∠AO2C is a right angle ∴ CO2 ∴. passing through the tangent point. Function: Use the radius of the tangent perpendicular to the tangent point to find the right angle or right triangle or chord tangent angle. Memory formula: to prove the tangent of the circle, the vertical radius passes through the outer end, and the straight line and the circle have * * * points, which proves that the radius is vertical, and the straight line and the circle are not given points, so the vertical line proves the radius. Example 5: In RT δ ABC, ∠ b = 90, the bisector of ∠A intersects BC at D, e is a point above AB, with D as the center and DB length as the radius. Proof: AC is the tangent of ⊙ D Enlightenment: There is no point on the circle, and the tangent can be obtained by proving the radius vertically. [canonical solution] let point d be DF⊥AC in f, ∫∠b = 90∴db⊥ab. And ∵AD is the angular bisector df ∵ AC ∴ db = df of ∠BAC. ∵DB is the radius of ⊙D, and ∴DF is also the radius of ⊙ d, so AC is the tangent of ⊙ D. (4) When two circles intersect, the auxiliary lines often cited are: male * * * chord; Lian Xin line function: ① bisect the common chord vertically with connecting lines; (2) Make it have rounded corners on the arc or form a quadrilateral inscribed in the circle to communicate the relationship between the two circles. Example 6: As shown in the figure, two circles intersect at B and C, AC cuts a small circle at C, ABE cuts a small circle at E, and even CE cuts a big circle at D ... Proof: AC = AD. Revelation: Since AC and AD form a triangle, it is only necessary to prove that ∠ ACD = ∠ ADC. But because these two angles are the peripheral angles of the great circle. So we should seek their relationship with the small circle. Observing the graph, we can find that ∠ CDA = ∠ E+∠ DAE. In this way, the problem becomes a problem about the angle of two circles, so it is necessary to make a chord and solve the problem with the help of the theorem of circumferential angle and tangent angle. Example 7: It is known that ⊙O 1 and ⊙O2 intersect at A and B, and the straight lines passing through A intersect at C and D respectively, connecting BO 1, BC, BO2 and BD. Verification: ∠CBD=∠O 1BO2 Enlightenment: In the graph where two circles intersect, the chord is an important auxiliary line. Because of the chord, the angles of these two circles are related in quantity. In other words, the outer angle after chord connection is equal to the inner diagonal or middle angle. In addition, two circles intersect and the connecting line bisects the chord vertically, which can enrich the known conditions. (5) When there is a common tangent of two circles, the frequently cited auxiliary line is a right triangle with the center distance as the hypotenuse and the sum (or difference) of the length and radius of the common tangent of two circles as the right length. As shown in the figure. Function: Use Pythagorean theorem or trigonometric function to calculate related quantity. (6) When two circles (or multiple circles) are tangent, the frequently cited auxiliary lines are: the tangent point leads to the common tangent of the two circles; Make a line connecting two circles. As shown in the figure (1), (2) and (3). Function: Connect the angle of the circle with the angle of the tangent of the chord, and connect the relationship between the two circles. Example 8: As shown in the figure, ⊙O 1 and ⊙O2 are tangent to A, BC is the common tangent of ⊙O 1 and ⊙O2, and B and C are tangent points. (1) Verification: AB ⊥ AC; (II) If r 1 and r2 are ⊙O 1 and ⊙O2, respectively, r 1=2r2, find the value of ■. Tips for auxiliary lines: memory formula: if you encounter circles and circles, it is very important to find the right position. The tangency of two circles is a common tangent, and the intersection of two circles is a common chord.