AB is the diameter of circle O, AC is the chord, OD ⊥ AC is at D, and the tangent AP of circle O is made through a. the extension line of AP and OD intersects P, connecting PC and BC (1) (2) prove that PC is tangent of circle o

(1) Obviously, OD = 1 / 2BC and OD ‖ BC
Because od ⊥ AC, ﹥ D is the midpoint of AC, and O is the midpoint of AB, so od is the median line of ⊥ ABC
(2) Link OC
In △ AOC, OA = OC, OD ⊥ AC
∴∠AOD＝∠COD
OA = OC, Op = Op
∴△AOP≌△COP
∴∠PCO＝∠PAO＝90°
⊙ PC is the tangent of ⊙ o

The diameter of ⊙ o middle chord ab ⊥ CD is known. The perpendicular foot is point F and the point E is on ab. EA = EC. Verification: AC * AC = AE * ab

Connecting CB
Then the triangle EAC and triangle cab are isosceles triangles, and the angle EAC is equal, so they are similar
AE:AC=AC:AB
AC*AC=AE*AB
The proof is complete

The chord ab of circle O is perpendicular to the diameter CD on F, E on AB [1] find AC square = AE * AB [2] extend EC to P, connect Pb = PE, try to judge the position relationship between Pb and circle o

Because the chord AB is vertical diameter CD, so AC = CB, so the angle cab = angle CBA because EA = EC, so the angle EAC = angle ace, so the isosceles triangle ace is similar to the isosceles triangle type ABC, so AC: EC = AB: AC = AC = EC * AB because EA = EC, AC = EA * AB, so the angle cob = 2 times angle CAE because of isosceles triangle type

In a circle of 50cm in diameter, if the chord AB = 40cm, the chord CD = 48CM and AB / / CD, then the distance between AB and AC is

22 or 8
When I was in high school, the teacher said that this kind of question was a free sub question

Known: radius of ⊙ o is 25cm, chord AB = 40cm, chord CD = 48CM, ab ∥ CD. Find the distance between the two parallel strings AB and CD

(1) As shown in Figure 1, connect ob, OD, and do om ⊥ AB to point n,
∵AB∥CD，
∴ON⊥CD，
∵AB=40cm，CD=48cm，
∴BM=20cm，DN=24cm，
The radius of ⊙ o is 25cm,
∴OB=OD=25cm，
∴OM=15cm，ON=7cm，
∵MN=OM-ON，
∴MN=8cm，
(2) As shown in Fig. 2, connect OB and OD, make a straight line om ⊥ AB, and cross CD to point n,
∵AB∥CD，
∴ON⊥CD，
∵AB=40cm，CD=48cm，
∴BM=20cm，DN=24cm，
The radius of ⊙ o is 25cm,
∴OB=OD=25cm，
∴OM=15cm，ON=7cm，
∵MN=OM+ON，
∴MN=22cm．
The distance between parallel strings AB and CD is 8cm or 22cm

The radius of circle O is 25 mm, AB, CD are two chords of circle O, and ab ∥ CD, ab = 40cm, CD = 48CM. Find the distance between AB and CD There is no picture

Draw EF ⊥ AB through point O, perpendicular foot is e, intersect CD with F
Connect ob, OD
∴BE=½AB=20
FD=½CD=24
OE＝√﹙OB²－BE²﹚＝15
OF=√﹙OD²－DF²﹚＝7
When AB and CD are in the same half circle, EF = oe-of = 8
When AB and CD are in different semicircles, EF = OE + of = 22
The distance between AB and CD is 8 or 22

Known: radius of ⊙ o is 25cm, chord AB = 40cm, chord CD = 48CM, ab ∥ CD. Find the distance between the two parallel strings AB and CD

0

As shown in the figure,
When AB and CD are on one side of the diameter,
In RT △ AOF, OA = 25cm, AF = 20cm,
∴OF=15cm．
Similarly, OE = 7cm,
The distance between the parallel line AB and CD is 15-7 = 8cm;
When AB and CD are not on the same side of diameter, the distance is 15 + 7 = 22cm
Therefore, D

As shown in the figure, AB is the diameter of ⊙ and the chord CD is vertically bisected ob, then ∠ BDC = () A. 15° B. 20° C. 30° D. 45°

Connect OC, BC
∵ chord CD vertical bisection ob
∴OC=BC
∵OC=OB
The △ OCB is an equilateral triangle
∴∠COB=60°
∴∠D=30°．
Therefore, C

As shown in the figure, the radius of circle O is 6cm, the chord AB is perpendicular to the diameter CD, and the length of chord AB is calculated by dividing CD into 1:3 parts

CD = 6 + 6 = 12cm
Let AB and CD intersect at point P and CP: PD = 1:3
CP = 1 / 4 × 12 = 3cm, PD = 12-3 = 9cm
Connecting AC and ad
CD is the diameter, so the angle CAD is a right angle
AB vertical CD
So AP 2 = Cp * PD = 3 × 9 = 27
AP = 3 √ 3cm
AB = 2AP = 6 √ 3cm

AB is the diameter of circle O, AC is the chord, OD ⊥ AC is at D, and the tangent AP of circle O is made through a. the extension line of AP and OD intersects P, connecting PC and BC (1) (2) prove that PC is tangent of circle o