As shown in the figure, the radius of ⊙ o is known to be 1, De is the diameter of ⊙ o, the tangent of ⊙ o is ad, C is the midpoint of AD, AE intersects ⊙ o at point B, and the quadrilateral bcoe is a parallelogram (1) Find the length of AD; (2) Is BC tangent to ⊙ o? If so, give proof; if not, give reasons

(1) Connecting BD, ∵ De is the diameter  DBE = 90 °,
∵ a quadrilateral bcoe is a parallelogram,
∴BC∥OE，BC=OE=1，
In RT △ abd, C is the midpoint of AD,
∴BC=1
2AD=1，
Ad = 2;
(2) Yes, for the following reasons:
As shown in the figure, connect ob. ∵ BC ∥ OD, BC = OD,
The quadrilateral bcdo is a parallelogram,
∵ ad is the tangent of circle o,
∴OD⊥AD，
The quadrilateral bcdo is rectangular,
∴OB⊥BC，
Then BC is the tangent of circle o

As shown in the figure, BD is the diameter of ⊙ o, and a is the diameter of ⊙ o At the midpoint of BC, a intersects BC at point E, passes through D as tangent line of ⊙ o, and extends line of BC at F, (1) Results: DF = EF; (2) AE = 2, de = 4, find DB length

(1) Connect OA, ∵ A is the midpoint of BC, ∵ OA ⊥ BC, ? OAE + ∠ AEG = 90 °, ?∵ AEG = ∠ fed, ? OAE + ∠ fed = 90 °, ∵ De is the tangent of the circle, ∵ de ⊥ BD, i.e., ∵ FDE + ⊥ ADB = 90 °, ∵ OA = OD, ﹤ OAE = ∠ ADB, ? DF = EF; (2) connect AB

As shown in the figure, AB is the diameter of circle O, ad is the chord, ∠ DBC = ∠ a (1) verification: BC is tangent to circle O (2) if OC is the vertical bisector of BD, the perpendicular foot is e, BD = 6 (2) If OC is the vertical bisector of BD, the perpendicular foot is e, BD = 6, CE = 4, find the length of AD. the first question is not required, but the second question is mainly

(2) ∵ be = BD = 3, ∵ be = BD = 3, ∵ be ⊙ be ⊸ OC, ∵ be ? OC, ∵ BeO = ∵ BEC = 90, ∵ EOB + ∵ OBE = 90 ∵ OBE + ∵ EBC = 90 ⊙ OBE ∵ EBC = 90 ? OBE ? be = BD = 3, ? be ? be ? OC, 8787878757\\\878787878787878757; be = median line

AB is the radius of semicircle o, the radius OC ⊥ AB, e is a point on ob, the chord ad ⊥ CE, and the perpendicular foot is g. conjecture the quantitative relationship between OE and of

Where is e? Any point of the extension of the ray PD? Point P and ab are on the same line? If my guess is true, then: in the right triangle ADB and triangle PDB, because the angle DAB + angle DBA = 90 degrees and the angle PDA = angle PBD, the angle PDA + angle DAB = 90 degrees

If OA = ob = OC = AB = 1, e, f are the midpoint of AB and OC, try to find the angle between OE and BF

By doing FG ‖ OE, we can get FG = 1 / 2oe = √ 3 / 4, BF = √ 3 / 2, BG = √ 7 / 4
The angle GFB = arccos2 / 3 can be obtained from cosine theorem

As shown in the figure, the radius OA of the circle center O is perpendicular to the chord BC, and ad = 2 BC = 8, find the radius of the circle center o

If it is, the solution is as follows: let the radius of circle o be r, then OA = ob = R. because the radius OA is perpendicular to the chord BC, so: according to the vertical diameter theorem, OA bisects chord BC, that is BD = BC / 2 = 4, then in RT △ OBD, OD = oa-ad = R-2, from the Pythagorean theorem, it can be concluded that the radius OA is perpendicular to the chord BC

As shown in the figure, in ⊙ o with radius r, chord AB = R, chord BC / / OA, then OA=

A is the point on the circle, O is the center of the circle, and OA is the radius R

As shown in the figure, AC, BD intersect at o point, and OA = OC, OB = OD

1、∵OA=OC,OB=OD,∠AOB=∠COD
∴△AOB≌△COD
∴AB=CD,∠ABO=∠CDO
∴AB∥CD
ABCD is a parallelogram
∴AD∥BC
2、∵OA=OC,OB=OD
ABCD is a parallelogram
∴AB∥CD,AD∥CB

As shown in the figure, ad and BC intersect at point O, ab ∥ CD, and OC = OD

∵AD∥BC,
∴∠A=∠D,∠C=∠B
∵OC=OD,
∴∠C=∠D,
∴∠A=∠B,
∴OA=OB