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As shown in the figure, △ ABC, ab = AC, e is a point on AB, f is a point on AC extension line, and be = CF, if EF and BC intersect at D, prove: de = DF
It is proved that FH ∥ AB cross BC extension line to H,
∵FH∥AB，
∴∠FHC=∠B．
And ∵ AB = AC,
∴∠B=∠ACB．
And ∵ ACB = ∵ FCH,
∴∠FHE=∠FCH．
∴CF=HF．
And ∵ be = CF,
∴HF=BE．
And ∵ FH ∥ ab,
∴∠BED=∠HFD，
And △ in hedbe,
∠B＝∠FHC
BE＝HF
∠BED＝∠HFD ，
∴△DBE≌△FHE（ASA）．
∴DE=DF．
CD is the chord of circle O. take CE = DF on CD, connect OE and of, and extend the intersection circle O at points a and B (1) Try to judge the shape of △ OEF and explain the reasons. (2) it is proved that arc AC = arc BD
1) OC = OD, triangle OCD is isosceles triangle, angle OCD = angle ODC; and because CE = Fe, so triangle OCE and triangle ODF are congruent, so OE = of
So the triangle OEF is also an isosceles triangle
2) Because the triangle OCE and the triangle ODF are congruent, the angle DOB = angle COA;
Two sides of the triangle BDO and the triangle COA are equal: od = ob = OA = OC and angle DOB = angle COA; therefore, the two triangles are congruent
So BD = AC
So the arc of BD and AC is AC = BD
Known: as shown in the figure, in the circle O, the chords AB = CD, e and F are the points on AB and CD respectively, and be = DF. Verification: OE = of
prove:
Make om ⊥ AB in M, on ⊥ CD in n
Then BM = 1 / 2 AB, DN = 1 / 2 CD [vertical diameter theorem]
∵AB=CD
∴BM=DN
Om = on
∵BE=DF
∴EM=FN
And ∵ ome = ∵ onf = 90
∴⊿OME≌⊿ONF（SAS）
∴OE=OF
26. As shown in the figure, in △ ABC, abbc intersects AC at point D, intersects with point E, passes through point D as DF ⊥ AC, and perpendicular foot is F. = AC, and circle O intersects with ab as diameter 26. As shown in the figure, in △ ABC, ab = AC, the circle O with ab as its diameter intersects BC at point D, AC at point E, passing through point D as DF ⊥ AC, and the perpendicular foot is f (1) Verification: DF is tangent line of ⊙ o;
(1) In RT triangle DFC, the angle FDC + angle FCD = 90 degrees connect OD, triangle BOD is isosceles triangle angle ODB = angle ABC, and triangle ABC is isosceles triangle angle ABC = angle FCD, so: angle ODB + angle FDC = angle FDC + angle FCD = 90 degrees angle ODF = 180 degrees - 90 degrees = 90 degrees DF is tangent of ⊙ o (2) when △ ABC is an equilateral triangle
As shown in the figure, in △ ABC, ab = AC, ⊙ o with ab as diameter intersects BC, AC at D and e respectively, passing through point D as DF ⊥ AC, and the perpendicular foot is f (1) Verification: DF is tangent of ⊙ o; (2) If AE= De, DF = 2, find the radius of ⊙ o
(1) Proof: connect OD, as shown in Fig,
∵AB=AC，
∴∠C=∠B，
∵OD=OB，
∴∠B=∠1，
∴∠C=∠1，
∴OD∥AC．
∴∠2=∠FDO，
∵DF⊥AC，
∴∠2=90°，
∴∠FDO=90°，
∵ od is the radius,
⊙ FD is the tangent line of ⊙ o;
(2) ∵ AB is the diameter of ⊙ o,
⊥ BC is ad ⊥ BC,
∵AC=AB，
∴∠3=∠4．
ν arc ed = arc DB
And arc AE = Arc de,
ν Arc de = arc DB = arc AE,
∴∠B=2∠4，
∴∠B=60°，
Ψ C = 60 °, and △ OBD is an equilateral triangle,
In RT △ CFD, DF = 2, ∠ CDF = 30 °,
∴CF=
Three
3DF=2
Three
3，
∴CD=2CF=4
Three
3，
∴DB=4
Three
3，
∴OB=DB=4
Three
3，
The radius of ⊙ o is 4
Three
3．
It is known that: as shown in the figure, ⊙ o with the diameter of AB on one side of the equilateral triangle ABC intersects with points D and e respectively, passing through point D as DF ⊥ BC, and the perpendicular foot is f (1) Verification: DF is tangent line of ⊙ o; (2) If the side length of the equilateral triangle ABC is 4, find the length of DF; (3) Find the area of the shadow part in the graph
It is proved that: (1) connect do
∵ △ ABC is an equilateral triangle,
∴∠A=∠C=60°．
∵OA=OD，
The △ oad is an equilateral triangle
∴∠ADO=60°，
∵DF⊥BC，
﹤ CDF = 90 ° - ∠ C = 30 °, (2 points)
∴∠FDO=180°-∠ADO-∠CDF=90°，
⊙ DF is the tangent line of ⊙ o; (3 points)
(2) ∵ △ oad is an equilateral triangle,
∴AD=AO=1
2AB=2．
∴CD=AC-AD=2．
In RT △ CDF,
∵∠CDF=30°，
∴CF=1
2CD=1．
∴DF=
CD2−CF2＝
3. (5 points)
(3) It can be seen from OE (2)
∴CF=1，
∴EF=1．
ν s right angled trapezoid FDOE = 1
2（EF+OD）•DF=3
Three
2，
The s sector OED = 60 π × 22
360=2π
3，
ν s shadow = s right angled trapezoid fdoe-s sector OED = 3
Three
2-2π
3. (7 points)
As shown in the figure, the chord AB and chord CD are perpendicular to e, and F is a point on ED, and CE = EF, extending AF intersects BD with H. it is proved that ah is perpendicular to BD
Proof: link AC
CE = EF, AE is the center line of △ ACF, and AE ⊥ CF is also high
So △ ACF is an isosceles triangle, ∠ ACF = ∠ AFC
Both ACF and B are arc ad, so ∠ ACF = ∠ B
Then ∠ ACF = ∠ B = ∠ AFC
∠B+∠BAH=∠AFC+∠BAH=90°
∴∠AHB=90° AH⊥BD
As shown in the figure, chord AB and chord CD are perpendicular to e, and F is a point on ED, and CE = EF, extending AF to intersect BD at h Confirmation: ah vertical BD
Without the graph, the position of a, B and D is uncertain, so it can't be solved
It is known that: as shown in the figure, AB is the diameter of ⊙ o, ad is the chord, OC is perpendicular to F, and ⊙ o is in E, connecting de and be, and ∠ C = ∠ bed (1) It is proved that AC is tangent of ⊙ o; (2) If OA = 10, ad = 16, find the length of AC
(1) It is proved that: ? bed = ∠ bad, C = ∠ bed, bad = ∠ C. (1 point) ? OC ⊥ ad at point F, ? bad + ∠ AOC = 90 °. (2 points) ? C + ∠ AOC = 90 °. [OA ⊥ AC. AC is tangent of ⊙ O. (4 points) (2) ? OC ⊥ ad at point F, ﹤ AF = 12ad = 8
As shown in the figure, it is known that AB is the diameter of circle O, BD is the tangent of circle O, the chord BC passing through point B is perpendicular to OD intersecting circle O at point C, and vertical is m. when BC is equal to BD and equal to 6cm, Calculate the area of the shadow part in the graph Half of the circle is in half, that is, half of the circle is shadow
Proof: connect OC
∵ OD ⊥ BC, O is the center of the circle,
The OD is equally divided into BC
∴DB=DC,
In △ OBD and △ OCD,
OB=OCDO=DODB=DC∴△OBD≌△OCD．（SSS）
∴∠OCD=∠OBD．
And ∵ AB is the diameter of ⊙ o, and BD is the tangent of ⊙ o,
⊙ o CD is the tangent of ⊙ o
∵ dB and DC are tangent lines, B and C are tangent points,
∴DB=DC．
DB = BC = 6,
The △ BCD is an equilateral triangle
∴∠BOC=360°-90°-90°-60°=120°,
∠OBM=90°-60°=30°,BM=3．
∴OM=BM•tan30°=3,OB=2OM=23．
The shaded part of S = s sector obc-s △ OBC
=120×π×(2
3) 2360-12×6×
Three
=4π-33（cm2）．
RELATED INFORMATIONS
- 1. As shown in the figure, the diameter ab of ⊙ o is perpendicular to the chord CD, and the plumb foot is e. if ﹤ cod = 120 ° and OE = 3 cm, then OD=______ Centimeter
- 2. The diameter of a circle has been increased by 5cm, the circumference has been increased by () and the area has been increased () Perimeter increased by () cm, area increased by () square centimeter?
- 3. As shown in the figure, O is a point on the diagonal of the square ABCD. The circle with o as the center and OA length as the radius is tangent to m and AB, and ad intersects EF respectively (2) If the side length of the square is 1, find the o radius of the circle
- 4. In the rectangular coordinate system, given the points a 5 to 2 and B 2 to 5, we can make a circle with point a as the center and ab as the radius, and judge the position relationship between the circle a and the X, Y axes
- 5. In △ ABC, if sin2a ≤ sin2b + sin2c sinbsinc, then the value range of a is () A. （0，π 6] B. [π 6，π） C. （0，π 3] D. [π 3，π）
- 6. If the derivative of function f (x) is f ˊ (x) = - SiNx, then the inclination angle of tangent line at point (π / 3, f (π / 3) is
- 7. The function y = 3sin (2x + π) 6) Monotone decreasing interval () A. [kπ−π 12，kπ+5π 12]（k∈Z） B. [kπ+5π 12，kπ+11π 12]（k∈Z） C. [kπ−π 3，kπ+π 6]（k∈Z） D. [kπ+π 6，kπ+2π 3]（k∈Z）
- 8. Let f (x) = cos (x + 2 π / 3) + 2cos 2 x / 2, X belongs to R If f (b) = 1, B = 1, C = radical 3, find a
- 9. The unit of the graph is shifted to the left by the unit of the graph
- 10. Simplification [2sin50 ° + sin80 ° (1 + √ 3tan10 °)] / cos5 °
- 11. As shown in the figure, ad is the angular bisector of △ ABC, de ⊥ AB, DF ⊥ AC, and the perpendicular feet are e and f respectively. If AB = 5, AC = 3, and the area of △ ABC is 16, find the length of de?
- 12. As shown in the figure, given that D is a point on the extension line of edge BC of △ ABC, DF ⊥ AB is at point F, intersects AC at point E, ∠ a = 40 ° and ∠ d = 30 °, then the degree of ∠ ACB is obtained______ Degree
- 13. 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
- 14. AB is the diameter of ⊙ o, CD, CB are the tangent lines of ⊙ o, B and D are tangent points, connecting OC to ⊙ o to e, straight line AE to BC, BD to F, G. verification: AF bisection ⊙ bad
- 15. As shown in the figure, circle O and circle 1 intersect at points a and B, AC is the diameter of circle O, and the extension lines of Ca and CB intersect circle O1 at points D and E, AC = 12, be = 30, BC = ad, Find ∠ C and de
- 16. It is known that AB and CD are the two diameters of ⊙ o, and the chord CE ∥ ab. it is proved that ad = AE
- 17. 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
- 18. As shown in the figure, in ⊙ o, D and E are points on the radius OA and ob respectively, and ad = be. Point C is a point on arc AB, connecting CD, CE, Co, ∠ AOC = ∠ BOC Confirmation: CD = CE
- 19. As shown in the figure, BD is the diameter of ⊙ o, point a is the midpoint of arc BC, ad intersects BC at point E, AE = 2, ed = 4 (1) The results were as follows: △ Abe ∽ abd; (2) Find the value of Tan ∠ ADB; (3) Extend BC to f and connect FD so that the area of △ BDF is equal to 8 3. Find the degree of ∠ EDF
- 20. As shown in the figure, in the rectangle ABCD, ab = 2, BC = 3, the vertical bisector of diagonal AC intersects ad respectively, BC connects CE at points E and F, then the length of CE is______ .