As shown in the figure, a is a point outside the circle O with radius 1, OA = 2, AB is the tangent line of ⊙ o, B is the tangent point, the chord BC ∥ OA, connecting AC, then the area of the shadow part is equal to () A. Three Four B. π Six C. π 6+ Three Eight D. π 4− Three Eight

Connect ob, OC,
∵ AB is the tangent line of the circle,
∴∠ABO=90°，
In the right angle △ ABO, OB = 1, OA = 2,
∴∠OAB=30°，∠AOB=60°，
∵OA∥BC，
Ψ cob = ∠ AOB = 60 ° and s shadow = s △ BOC,
△ BOC is an equilateral triangle with a side length of 1,
The shaded part of S = s △ BOC = 1
2×1×
Three
2=
Three
4．
Therefore, a

As shown in the figure, a is a point outside the circle O with radius 1, OA = 2, AB is the tangent line of ⊙ o, B is the tangent point, the chord BC ∥ OA, connecting AC, then the area of the shadow part is equal to () A. Three Four B. π Six C. π 6+ Three Eight D. π 4− Three Eight

0

According to the law of right triangle, AOB = 60 degrees, BC / / OA, BOC = 60 degrees
No map, shadow area you should be able to find

As shown in the figure, it is known that AB is the chord of ⊙ o, the radius OA = 20cm, ∠ AOB = 120 ° and the area of ⊙ AOB is obtained

The transition point O is OC ⊥ AB to C, as shown in the following figure:  AOC = 12 ﹤ AOB = 60 °, AC = BC = 12ab, ﹤ in RT △ AOC, ﹤ a = 30 ° OC = 12oa = 10cm, AC = oa2 − oc2 = 202 − 102 = 103 (CM), ab = 2Ac = 203cm ﹥ the area of AOB = 12ab · OC = 12 × 203 × 10 = 1003 (cm2)

As shown in the figure: isosceles △ ABC, take the waist AB as the diameter, make ⊙ o, the bottom edge BC is at P, PE ⊥ AC, and the perpendicular foot is e Verification: PE is tangent of ⊙ o

Proof: connect Op,
∵ AB is the diameter of ⊙ o,
∴∠APB=90°，
∵AB=AC，
∴BP=CP，
∵OB=OA，
∴OP∥AC，
∵PE⊥AC，
∴OP⊥PE，
∵ Po is the radius,
⊙ PE is the tangent of ⊙ o

As shown in the figure, in RT △ ABC, ∠ ACB = 90 °, BAC = 30 °, △ ADC and △ Abe are equilateral triangles, and de intersects AB at point F. it is proved that f is the midpoint of de

In RT △ ABC, ∠ ACB = 90 °. AC = BC, point D is a point in the triangle, and ∠ ADC = 135 ° it is proved that AB is the tangent of △ ADC circumcircle

It is proved that let the center of circumscribed circle of △ ADC be o, and connect AO and Co
∵∠ADC=135°
∴∠AOC=(180°-∠ADC)*2=90°
And AO = Co
∴∠OAC=∠ACO=45°
∴∠BAO=∠OAC+∠CAB=45°+45°=90°
/ / AB is the tangent of circle o
That is, AB is the tangent of the circumscribed circle of △ ADC

As shown in the figure, take one waist ab of the isosceles triangle ABC as the diameter of the circle O intersects the other waist at point E, and the disclosure side BC is verified at point D: BC = 2DE

Have you ever looked at it carefully? I asked you that the angle is equal: x0d congruence, triangle Dec is equal to BAC question: x0d triangle Dec is equal to BAC. How to answer: x0d have you seen my first process? Oh, I copied the answers to the questions you asked me. I was the first one to answer. I was really angry. I connected AD and be to prove that triangle abd and BDE are congruent, Then we get that AE is equal to BD, and then because the triangle ABC is an isosceles triangle, so EC = DC, so the triangle Dec is all equal to BAC. The question is: as you said, ED is parallel to AB, but it seems wrong to me. Look at my process and explain it to me. Excuse me. Answer: x0d halo, did you tell me which two sides of isosceles triangle are equal, Come on, my previous efforts are in vain, and they are all wrong. Take a look at it yourself. Ask: "x0d" the circle O with the diameter of AB at one waist of the isosceles triangle ABC intersects the other waist with point e "is very clear. But can you tell me now

The radius OA and ob of circle O intersect with chord CD at e and f respectively, and CE = cf. it is proved that OE = of; AC = BD

First question:
∵ OC = OD,  OCE ≌ ODF, CE = DF,  OCE ≌ △ ODF, ᙽ OE = of
Second question:
∵, ∵∵, ∵∵∵, ∵∵, ∵∵∵, ∵∵∵∵∵∵∵∵∵∵∵∵∵

As shown in the figure, in △ ABC, ab = AC, point D is on BC, de ‖ AC intersects point E, DF ‖ AB intersects AC at point F. verification: de + DF = ab

Maybe not the most standard solution
prove:
Because de ∥ AC, DF ∥ ab. (known)
Therefore, a quadrilateral AEDF is a parallelogram
So AF = ed, AE = DF
Because AB = AC, (known)
So ∠ B = ∠ C (equilateral and equal angle)
Because ∠ C + ∠ EDC = 180 ° (two lines are parallel, and the inner angle of the same side is complementary)
So ∠ B + ∠ EDC = 180 ° (equivalent substitution)
Because ∠ EDB + ∠ EDC = 180 ° (flat angle complementary)
So ∠ B = ∠ EDB (equivalent substitution)
So EB = ed
So de + DF = ab

As shown in the figure, a is a point outside the circle O with radius 1, OA = 2, AB is the tangent line of ⊙ o, B is the tangent point, the chord BC ∥ OA, connecting AC, then the area of the shadow part is equal to () A. Three Four B. π Six C. π 6+ Three Eight D. π 4− Three Eight