As shown in the figure, given that the radius of ⊙ o is 1, AB and ⊙ o are tangent to point a, OB and ⊙ o intersect at point C, CD ⊥ OA, and the perpendicular foot is D, then the value of cos ⊥ AOB is equal to () A. OD B. OA C. CD D. AB

As shown in the figure, given that the radius of ⊙ o is 1, AB and ⊙ o are tangent to point a, OB and ⊙ o intersect at point C, CD ⊥ OA, and the perpendicular foot is D, then the value of cos ⊥ AOB is equal to () A. OD B. OA C. CD D. AB

∵CD⊥OA，
∴∠CDO=90°，
∵OC=1，
∴cos∠AOB=OD：OC=OD．
Therefore, a

It is known that the radius of circle O is 1, AB and circle O are tangent to point a, OB and circle O intersect with C, CD ⊥ OA, and the perpendicular foot is D, cos ⊥ AOB=

According to the image:
cos∠AOB=OD/CD=OD：1=OD
So this problem is equal to OD

It is proved that AE = BF = CD

C. If D is the trisection point of arc AB, then ∠ AOC = ∠ cod = ∠ DOB = 30 °, AC = CD = dbao = ob, ∠ AOB = 90 °, then ∠ OAB = ∠ oba = 45 ° OA = OC, ∠ AOC = 30 ° then ∠ OAC = 75 ° and ∠ OAB = 45 ° then ∠ BAC = 30 °, ACO = ∠ Cao = 75 ° then ∠ AEC = 75 ° then ace is an isosceles triangle, AC =

As shown in the figure, AB, CD intersect at point O, AC ∥ BD. it is proved that OA? Od = ob? OC

Proof: ∵ AC ∥ BD, (1 point)
Ψ a = ∠ B, ∠ C = ∠ D. (2 points)
﹤ AOC ∽ BOD. (4 points)
∴OA
OB＝OC
OD (6 points)
/ / OA · od = ob · OC. (8 points)

As shown in the figure, AB, CD intersect at point O, AC ∥ BD. it is proved that OA? Od = ob? OC

Proof: ∵ AC ∥ BD, (1 point)
Ψ a = ∠ B, ∠ C = ∠ D. (2 points)
﹤ AOC ∽ BOD. (4 points)
∴OA
OB＝OC
OD (6 points)
/ / OA · od = ob · OC. (8 points)

As shown in the figure, O is the intersection point of diagonal AC and BD of rectangular ABCD, e, F, G, h are points on AO, Bo, CO and do respectively, and AE = BF = CG = DH Verification: quadrilateral efgh is rectangular

Prove that: ∵ quadrilateral ABCD is a rectangle,
/ / AC = BD; AO = Bo = co = do, (2 points)
∵AE=BF=CG=DH，
∴OE=OF=OG=OH，
The quadrilateral efgh is a parallelogram. (4 points)
∵ OE + og = fo + Oh, eg = FH,
The quadrilateral efgh is a rectangle (parallelogram with equal diagonals is a rectangle). (7 points)

The diagonals AC and BD of rectangular ABCD intersect at point O, points E and F are on OA and OD respectively, and AE = DF. It is proved that the quadrilateral ebcf is isosceles trapezoid First, we must prove that ebcf is trapezoid. Do not use parallel lines to divide segments in proportion. We did not learn to use auxiliary lines with two overlapping points

Proof: ABCD is a rectangle, so AC = BD
OA=AC/2,OD=BD/2
Therefore, OA = OD. ∠ oad = ∠ ODA = (180 - ∠ AOD) / 2
OE=OA-AE,OF=OD-DF
Because AE = DF, OE = of
Therefore ∠ OEF = ∠ ofe = (180 - ∠ AOD) / 2
So ∠ OEF = ∠ oad, EF ‖ ad ‖ BC
In △ OEB and △ OFC
OE=OF,
∠EOB=∠FOC,
OB=OC,
So △ OEB ≌ △ OFC
BE=CF
Because be is not parallel to CF, ebcf is trapezoidal
And be = CF, so it is isosceles trapezoid

It is known that, as shown in the figure, ⊙ 0 is made by taking the oblique edge ab of RT △ ABC as the diameter, and D is the point on BC, and there is an arc AC = arc CD, connected with CD and BD, take a point E on the extension line of BD, so that

(1) Proof: connect OC, ad,
A kind of
AC=
CD，
∴OC⊥AD，∠ADC=∠DBC，
While, if the option DCE= option CBD, then the option DCE= option ADC,
∴CE∥AD，
∴OC⊥CE，
CE is the tangent of ⊙ O;
(2) Let ad cross OC to point F,
∵ AB is the diameter,
∴∠ADB=90°，
By CE ∥ ad,
∴∠E=90°，
A kind of
AC=
CD，
∴OC⊥AD，AF=DF，
In RT △ CED, if de = x, CE = 2x and CD = 2
5，
According to Pythagorean theorem, x 2 + (2x) 2 = (2)
5)2，
The solution is: x = 2,
∴DE=2，CE=4，
∵∠E=∠OCD=∠ADE=90°，
The quadrilateral CEDF is a rectangle,
∴AF=DF=CE=4，CF=DE=2，
In RT △ oaf, let OA = R, according to Pythagorean theorem, R2 = 42 + (X-2) 2
∴r=5．
A: the radius is 5

As shown in the figure, in RT △ ABC, ∠ C = 90 °, a = 30 °, AC = 6cm, CD ⊥ AB in D, C as the center of the circle, CD as the radius of the arc, intersection BC with E, then the area of the shadow part in the figure is () A. （3 Two 3-3 4π）cm2 B. （3 Two 3-3 8π）cm2 C. （3 3-3 4π）cm2 D. （3 3-3 8π）cm2

∵∠A=30°，AC=6cm，CD⊥AB，
∴∠B=60°，∠BCD=30°，CD=3cm，BD=
3cm，
So s △ BDC = 1
2BD×DC=3
Three
2cm2, s sector CED = 30 π × 32
360=3π
4．
Therefore, the area of shadow is: (3)
Three
2-3π
4）cm2．
Therefore, a

It is known that AB is the diameter of circle O, CD is perpendicular to AB, AC arc is equal to FC arc, and AE = CE

Connect CO to AF and h to OE
AC arc is equal to FC arc
So C is the midpoint of the AF arc
Then OC ⊥ AF
Because CD ⊥ AB OC = OA ∠ cod = ∠ AOH
△COD≌△AOH
Then od = Oh
Then ch = ad
We can deduce △ ead ≌ △ EVH
AE=CE