Given the diameter ab of circle O, perpendicular to the chord CD at point E, connect CO and extend the intersection ad at point F. if CF is perpendicular to ad, ab = 2, find the length of CD process CF is not the center of the circle

Given the diameter ab of circle O, perpendicular to the chord CD at point E, connect CO and extend the intersection ad at point F. if CF is perpendicular to ad, ab = 2, find the length of CD process CF is not the center of the circle

First of all, AB is the diameter, CD is perpendicular to AB, which means that point E is the midpoint of CD, AB is perpendicular and bisects CD. Because CO is also the diameter, ad is perpendicular to F, then Co also bisects ad vertically. After connecting AC, we can find that because AE bisects CD vertically, so side AC = ad, similarly, CF bisects ad vertically, so AC = CD

As shown in the figure, it is known that AB is the chord of ⊙ o, OB = 2, ∠ B = 30 °, C is any point on the chord AB (not coincident with points a and b), connect CO and extend CO to intersect ⊙ o at point D, and connect ad. (1) chord ab=______ (2) when ∠ d = 20 °, calculate the degree of ∠ BOD

(1) As shown in the figure, O is taken as OE ⊥ AB in E, where ⊥ e is the midpoint of ab. in RT △ OEB, OB = 2, ⊥ B = 30 °, ⊥ OE = 1, ⊥ be = 3, ⊥ AB = 2be = 23; (2) solution 1: ⊥ BOD = ⊥ B + ⊥ BCO, ⊥ BCO = ⊥ a + ⊥ D. ⊥ BOD = ⊥ B + ⊥ a + ⊥ D. ⊥ (3 points) and ∵ BOD = 2 ∵ a, ∵ B = 30 °, ∵ d = 20 °, ∵ 2 ∵ a = ∵ B + ∵ a + ∵ d = ∵ a + 50 °, ∵ a = 50 ° (4 points) ≠ BOD = 2 ∠ a = 100 ° (5 points) solution 2: as shown in the figure, connect OA. ∵ OA = ob, OA = OD, ∵ Bao = ∵ B, ∵ Dao = ∵ D, ∵ DAB = ∵ Bao + ∵ Dao = ∵ B + ∵ D. & nbsp (3 points) and ∵ B = 30 °, d = 20 °, DAB = 50 ° and (4 points) ∠ BOD = 2 ∠ DAB = 100 ° (5 points)

C is a point on the diameter ab of the circle O, and the chord De is made through the point C so that CD = Co. if the arc ad is 30 degrees, the size of the arc be is calculated

Connecting od and BD, because the angle AOD is 30, OC = CD, so cod isosceles triangle, angle cod = angle CDO = 30, so angle BOD = 120, because ob = OD. So angle OBD = angle ODB = 30, so angle EDB = angle Edo + angle ODB = 30 + 30 = 60
Because the circumference angle is half of the center angle, the angle AOE = 120, so the arc be = 120

As shown in the figure, ab ⊥ CD in B, CF intersection AB in E, CE = ad, be = BD, verification: CF ⊥ ad

It is proved that in RT △ CBE and RT △ abd, CE = ADBE = BD, ≌ RT △ CBE ≌ RT △ abd & nbsp; (HL), ∤ C = ∨ a, ∨ AEF = ∨ CEB, ∨ CBE = ∨ AFE = 90 ° and ⊥ CF ⊥ ad