As shown in the figure, AB is the diameter of ⊙ o, CD is the chord, CE ⊥ CD intersects AB with E, DF ⊥ CD intersects AB with F, proving that AE = BF

It is proved that if O is used as og ⊥ CD, according to the vertical diameter theorem, og is vertically divided into CD, then CG = DG,
∵CE⊥CD，DF⊥CD，OG⊥CD，
∴CE∥OG∥DF，
∵CG=DG，
∴OE=OF，
∵OA=OB，
∴AE=BF．

In ⊙ o, AB is diameter, CD is chord, CE ⊥ CD is at point C, crossing AB at point E, DF ⊥ CD at point D, crossing AB at point F. verification: AE = BF Set M as the point of connection, Then OM is perpendicular to CD Because CE is perpendicular to CD, DF is perpendicular to CD So CE is parallel to OM and parallel to DF (in the same plane, two lines perpendicular to the same line are parallel to each other) Because m is the midpoint of CD (set) So OE = of (parallel line segments are proportional) And OA = ob (radius inside the circle) So oa-oe = ob-of AE = BF So OE = of (parallel line segments are proportional) I can't understand this step Did not learn parallel line, line segment proportional Have you proved this problem with the method you have learned in junior high school textbooks?

Is the trapezoid median line theorem, also known as the parallel line equidistant line theorem, this in the junior high school textbook is deleted
It means that between several parallel lines, any line segment is equally divided. The most typical example is the grid of exercise books. You take a ruler and let one side of the ruler be bisected by the grid line. Then you turn the ruler. At this time, the edge of the oblique ruler is equally divided as the straight edge

AB is the diameter of circle O, CD is the chord of circle O, AE is perpendicular to CD, BF is perpendicular to CD, e and F are perpendicular feet, indicating CE = DF

Because AE, BF and og are perpendicular to CD, AE ∥ BF ∥ og. Known, OA = ob, we can get: Ge = GF. (different lines are cut by the same group of parallel lines in proportion) (another explanation: Ge and gf are equal to the distance from point O to AE and BF respectively, OA = o

In the circle O, AB is the diameter, CD is the chord intersecting AB, and ab = 10, CD = 8, AE is perpendicular to CD to e, BF is perpendicular to CD to F, and ae-bf =?

Ae-bf = 6, let AB and CD intersect at point G, connect center O of circle O and midpoint h of chord CD, take point m on line segment AG, make GM = GB, make Mn ∥ BF, MP ∥ CD, intersect CD and AE at N and P ∵ Mn ∵ BF ∵ NMG = ∵ GBF ∵ GM = GB ∵ MGN = ≌ △ BGF  Mn = BF

It is known that the chord CD of the center O is perpendicular to f and the diameter AB is perpendicular to F, e is on CD, AE = CE, 1) the square of Ca = AE × CD 2) given CA = 5, EA is equal to 3, find sin ∠ EAF 2012 Ya'an

1. It can be proved that isosceles △ ACD is similar to isosceles △ AEC, so there is AC / CE = DC / AC, so the square of AC = CE * CD, AE = CE, so the square of AC = AE * CD,
2, because the square of AC = AE * CD, it is calculated that CD = 25 / 3, CF = 1 / 2 * CD = 25 / 6, EF = cf-ce = 7 / 6,
sin∠EAF=EF/AE=7/18

As shown in the figure, AB is the diameter of ⊙ o, and CD is the chord. The vertical lines of CD are drawn through two points a and B respectively, and the vertical feet are e and F. verification: EC = DF

It is proved that O is om ⊥ CD is at point M,
∵OM⊥CD，
∴CM=DM，
∵AE⊥EF，OM⊥EF，BF⊥EF，
∴AE∥OM∥BF，
∵ AB is the diameter of ⊙ o,
∴OA=OB，
/ / OM is the median line of trapezoidal aefb,
∴EM=FM
/ / em-cm = fm-dm, i.e., EC = DF

As shown in the figure: AB and AC are the two chords of ⊙ o, extend CA to point D, make ad = AB, if ∠ d = 40 °, calculate the degree of ⊙ BOC

∵AD=AB，∠D=40°，
∴∠ABD=∠D=40°，
∴∠BAC=∠ABD+∠D=80°，
∴∠BOC=2∠BAC=160°．

As shown in the figure, AB and AC are the two chords of the circle center O. extend CA to point d so that ad = ab. if ∠ ADB = 30 °, then ∠ BOC=

30 degrees. The center angle is half of the degree of the opposite circle angle

As shown in the figure, AB is the diameter of ⊙ o, am, BN cut ⊙ o at point a, B, CD, am, BN at point D, C, do bisection ⊙ ADC (1) It is proved that CD is tangent of ⊙ o; (2) If ad = 4, BC = 9, find the radius r of ⊙ o

(1) It is proved that OE ⊥ CD is made at point e through point o,
∵ am cut ⊙ o at point a,
∴OA⊥AD，
And ∵ do bisection ∵ ADC,
∴OE=OA，
∵ OA is the radius of ⊙ o,
⊥ o e is the radius of ⊙ o, and OE ⊥ DC,
⊙ CD is the tangent of ⊙ o
(2) Pass point D as DF ⊥ BC at point F,
∵ am, BN cut ⊙ o at points a and B respectively,
∴AB⊥AD，AB⊥BC，
The quadrilateral ABFD is a rectangle,
∴AD=BF，AB=DF，
And ∵ ad = 4, BC = 9,
∴FC=9-4=5，
∵ am, BN, DC cut ⊙ o at points a, B, e respectively,
∴DA=DE，CB=CE，
∴DC=AD+BC=4+9=13，
In RT △ DFC, DC2 = df2 + FC2,
∴DF=
DC2−FC2＝
132−52=12，
∴AB=12，
The radius r of ⊙ o is 6

Am is the diameter of circle O. a point B on the circle O is perpendicular to am, and its extension line intersects circle O with C chord, CD with am with ECD and ab with FCD = AB, which means CE = EF * ed If the extension line of CD AB intersects point F, and CD = AB, then whether the conclusion is true, please prove I've proved the first question,

Simple
Connect EB BD
Angle EBA = angle ACE (angle ACB = angle ABC, angle ECB = angle EBC)
The angle ace = angle cab (CD = AC = AB)
Angle cab = angle CDB (same chord)
therefore
Angle EBA = angle CDB, angle DEB is common angle
Triangle EBD is similar to EFB
EB/EF=ED/EB
And EB = EC
Replace and get what you want

As shown in the figure, AB is the diameter of ⊙ o, CD is the chord, CE ⊥ CD intersects AB with E, DF ⊥ CD intersects AB with F, proving that AE = BF