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As shown in the figure, in the circle inscribed square ABCD with the side length of 1, P is the midpoint of the edge CD, and the straight line AP intersects the circle and the point e to find the length of the chord De
Let: the center of the circle is O, connect OD, OE, and make the chord center distance of De of ∠ DAE = ∠ DOE / 2 = ∠ DOF (the angle of the center of the circle opposite to the same arc is twice of the circumference angle) / / RT Δ DOF ∷ of / DF = ad / PD = 2 / 1 = = > of = 2DF ﹥ DF ∽ DF ∽ DF ∽ DF = ad / PD = 2 / 1 = = > of = 2DF ﹥ DF ﹥ DF ∧ (2DF) 2 ﹣ OD ﹤ 5; 5; 2 / 2) = = = = > D
If the radius of ⊙ o is 12 (1) Find the length of string Mn; (2) Connect OM and on to find the degree of the center angle ∠ mon
(1) ?????????????????
As shown in the figure, the edge ab of the square ABCD with side length of 1 is the diameter of ⊙ o, CF cuts ⊙ o at point E, intersects ad at point F, and connects be (1) Calculate the area of △ CDF; (2) Find the length of segment be
(1) Da, CB and CF are tangent lines of ⊙ o,
∴AF=EF，CE=CB．
Let AF = x, then in RT △ FDC, (1-x) 2 + 1 = (x + 1) 2,
∴x＝1
4．
∴S△FDC=1
2×CD×DF＝3
8．
(2) Connect OC to be at point G and connect OE
∵ CE, CB are tangent lines of ⊙ o,
∴CE=CB．
And ∵ OE = ob,
﹣ CO is vertically bisected into be
In RT △ OBC, OC =
BC2+OB2＝
Five
2．
∵S△BOC=1
2×OB×BC＝1
2×BG×OC，
∴BG=
Five
5，
∴BE=2BG=2
Five
5．
As shown in the figure, the side length of the square ABCD is 4, the diameter of BC is taken as the circle, the tangent line of the circle is made through point a, the intersection of DC and E, and the tangent point is f (1) The area of ade is calculated (2) Find the length of BF
(1) ∵ ab ⊥ BC, ᙽ AB is tangent of circle O, and AE is tangent of circle O, ᙽ AB = AF = 4. Similarly, EF = EC is obtained. If EF = EC = x, then de = dc-ec = 4-x, AE = AF + EF = 4 + X. in RT △ ade, by using Pythagorean theorem, AE2 = ad2 + de2, that is (4 + x) 2 = 42 + (4-x) 2, the solution is: x = 1, ᚉ de = 4 + X
As shown in the figure, the quadrilateral ABCD is inscribed in ⊙ o, AB= It is proved that AB2 = be · CD
Proof: connect AC,
∵ EA cut ⊙ o in a,
∴∠EAB=∠ACB．
A kind of
AB=
AD，
∴∠ACD=∠ACB，AB=AD．
So ∠ EAB = ∠ ACD
The quadrilateral ABCD is inscribed in ⊙ o,
∴∠ABE=∠D．
∴△ABE∽△CDA．
So ab
CD=BE
Da is ab · Da = be · CD
∴AB2=BE•CD．
AB is the diameter of circle O, ad is the chord, angle DAB + 22.5 ^, extend AB to point C such that angle ABCD = 45 ^, find CD is tangent of circle o
(1) It is proved that: connected do, ∵ Ao = do, Dao = ∠ ADO = 22.5 °. doc = 45 °. And ?∵ ACD = 2 ∠ DAB, ACD = doc = 45 °. ? ODC = 90 °. CD is tangent of ⊙ o
AB is the diameter of the circle O, AC is the chord, the bisector ad of ∠ BAC intersects the circle O at point D, de ⊥ AC, the extension of AC at point E and OE at point F Prove that De is tangent of circle O
Connecting OD, then △ AOD is an isosceles triangle, ∠ ADO = ∠ Dao. From the bisector of ∠ BAC, we know ∠ Dao = ∠ DAC, so ∠ ADO = ∠ DAC, OD ‖ AC. because de ⊥ AC, de ⊥ OD, so de ⊥ od is a tangent line of circle o
AB is the diameter of the center O, AC is the chord, the bisector ad of ⊥ BAC intersects the center of circle O at point D, the extension of AC intersects AC at point E, and OE intersects ad at point F Proof: De is tangent line of center o
Connect OD
∵AD=OA
∴∠ODA=∠OAD
And ? oad = CAD
∴∠ODA=∠CAD
∴OD‖AC
∵DE⊥AC
∴DE⊥OD
⊙ De is the tangent of ⊙ o
Elective course 4-1: geometry proof As shown in the figure, AB is the diameter of ⊙ o, AC is the chord, the bisector ad of ⊙ BAC intersects ⊙ o at point D, de ⊥ AC, and the extension line of AC intersects at point E. OE intersects ad at point F (1) It is proved that De is tangent of ⊙ o; (2) If AC AB＝3 5, find AF DF value
(1) It is proved that by connecting OD, ∠ ODA = ∠ oad = ∠ DAC (2) od ∥ AE, AE ⊥ De (3 points) de ⊥ OD, and OD is the radius ⊥ De is ⊙ o tangent line (5 points) (2) if D is DH ⊥ AB in H, then ∠ DOH = ∠ cabcos ∠ DOH = cos ∠ cab = acab = 35 (6 points
As shown in the figure, in △ ABC, ∠ C = 90 °, AC = BC, ad bisection ∠ cab intersects BC at point D, de ⊥ AB, the foot perpendicularly is e, and ab = 6cm, then the circumference of △ DEB is () A. 4cm B. 6cm C. 8cm D. 10cm
∵ ad bisection ∠ cab intersects BC at point D
∴∠CAD=∠EAD
∵DE⊥AB
∴∠AED=∠C=90
∵AD=AD
∴△ACD≌△AED．（AAS）
∴AC=AE，CD=DE
∵∠C=90°，AC=BC
∴∠B=45°
∴DE=BE
∵AC=BC，AB=6cm，
ν 2bc2 = AB2, that is BC=
AB2
2=
Sixty-two
2=3
2，
∴BE=AB-AE=AB-AC=6-3
2，
∴BC+BE=3
2+6-3
2=6cm，
The circumference of DEB = de + DB + be = BC + be = 6 (CM)
Another method: after proving the congruence of triangles,
∴AC=AE，CD=DE．
∵AC=BC，
∴BC=AE．
The circumference of △ DEB = DB + de + EB = DB + CD + EB = CB + be = AE + be = 6cm
Therefore, B
RELATED INFORMATIONS
- 1. As shown in the figure, PA and Pb are two tangent lines of circle O, a and B are tangent points, AC is the diameter of circle O, ∠ BAC = 35 ° and the degree of ∠ P is calculated
- 2. If the length of a string is equal to the radius, then the radians of the central angle of the chord are? Why?
- 3. In a circle, if the chord AB = 1 / 3 of the root of 2 times and the chord center distance is 1, what is the radius of the circle? A, 2 B, 13 C, 11, D, 3 What to choose?
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- 6. It is known that the radii of circle O1 and circle O2 are 3cm and 5cm respectively, and if they are inscribed, the position relationship between the center of circle and O1O2? What is the positional relation? Which is the extrinsic intersection and tangent? It is known that the radius of circle O1 and circle O2 are 3cm and 5cm respectively, and if they are inscribed, the center distance of circle is 5cm!
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- 8. As shown in the figure, point h is the vertical center of △ ABC, the circumscribed circle ⊙ O2 of ⊙ O1 and ⊙ BCH with ab as diameter intersect at point D, and extend ad to intersect ch at point P, Verification: point P is the midpoint of Ch
- 9. Circle O1 and circle O2 intersect at two points a and B, P is the point on circle O1, connecting PA and Pb, intersecting circle O2 with C and D. It is proved that po1 is perpendicular to CD
- 10. As shown in the figure, ad is the diameter of ⊙ o, AC is the chord, ∠ CAD = 30 °, ob ⊥ ad is in O, AC is in B, ab = 5, and the length of BC is obtained
- 11. Given that AB is the diameter of ⊙ o, ab = 16, P is the midpoint of ob, the chord CD passes through the point P, ∠ APC = 30 °, then what is CD?
- 12. As shown in the figure, the diameter ab of ⊙ o is perpendicular to the chord CD at m, and M is the midpoint of radius ob, CD = 8cm
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- 14. AB is the diameter of circle O, the chord CD is perpendicular to e, P is on the circle, ∠ CBP = ∠ C, find CB / / PD
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- 16. As shown in the figure, ad is the middle line of △ ABC, be intersects AC to e, and ad to F, and AE = EF. Verification: AC = BF
- 17. The ad of triangle ABC and CB is equal to c.ca CB.CE be equal to CD.AE Intersect BD with F, if the angle AEC is equal to 67 degrees
- 18. If the diameter ab of O intersects with the chord CD at point E, given AE = 6cm, EB = 2cm, ∠ CEA = 30 °, then the length of chord CD is () A. 8cm B. 4cm C. 2 Fifteen D. 2 Seventeen
- 19. As shown in the figure, AB and CD intersect at the point O, O is the midpoint of AB, ∠ a = ∠ B, is Δ AOC and Δ BOD congruent? Why?
- 20. As shown in the figure, AB and CD are the two chords of circle O. extending Ba and DC respectively, intersecting at points P, m, n are the midpoint of arc AB and arc CD, and Mn ⊥ Po Speed, please