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It is known that: as shown in the figure, the diameter EF of ⊙ o intersects the chords AB, CD at the points g and h respectively, and Ag = BG, ch = DH
Connect OA ob, so the triangle OAB is isosceles triangle, and Ag = BG, so AB is vertical EF, similarly CD is vertical EF, so AB / / CD is vertical
As shown in the figure, the diameter ab of ⊙ o intersects with chord EF at point P, and the intersection angle is 45 degrees. If Pe2 + PF2 = 8, then AB is equal to______ .
Make og ⊥ EF to g, connect OE,
According to the vertical diameter theorem, if eg = FG = x, then PE = x + PG, PF = x-pg,
And ∵ Pe2 + PF2 = 8,
∴（x+PG）2+（x-PG）2=8，
The results show that 2x2 + 2pg2 = 8, X2 + PG2 = 4,
∵ the angle of intersection is 45 degrees,
∴OG=PG，
∴OE2=OG2+EG2=4，
That is, the radius of the circle is 2,
The diameter is 4
As shown in the figure, AB, CD are the two chords of circle O with radius 5, ab = 8, CD = 6, Mn is the diameter, ab ⊥ Mn, CD ⊥ Mn, P is any point on EF What is the minimum value of PA + PC
When the intersection point of BC and EF is p, PA + PC is the shortest
Linking OA, OC, from Pythagorean theorem
OE=3,OF=4
∴EF=7
∵AB‖CD
∴BE/CF=EP/PF
4/3=EP/PF
EP+PF=7
∴EP=4,PF=3
∴BP=4√2,PC=3√2
The shortest distance of PA + PC = BC = 7 √ 2
As shown in the figure, in △ ABC, point D is a point on AB, passing point D is de ∥ BC, intersection edge AC is at point E, crossing point E is EF ∥ DC and crossing ad is at point F. ad is known= (1) the value of AE vs. AC (2) the value of AF vs. AB (hint: the segments of parallel lines in the third day of junior high school are proportional and similar triangles)
Because of De / / BC, AE / AC = ad / AB = 2 roots (6) / 8 = roots (6) / 4
Because of Fe / / DC, AF / ad = AE / AC = radical (6) / 4
AF / AB = (AF / AD) * (AD / AB) = (root (6) / 4) * (2 root (6) / 8) = 3 / 8
Let ABC be an equilateral graph where △ DC = 60 ° respectively 1. Verification: quadrilateral cdef is parallelogram 2. If BF = EF, verification: AE = ad
∵ △ ABC is an equilateral triangle
∴∠B=60°
∵∠EFB=60°
∴∠B=∠EFB
EF ‖ BC
∵ DC=EF
The quadrilateral efcd is a parallelogram (a set of opposite sides parallel and equal)
(2) Connect be,
∵∠EFB=60°,BF=EF
The triangle bef is an equilateral triangle
That is be = BF = EF, ∠ Abe = 60 °
∵ a quadrilateral efcd is a parallelogram
∴CD=EF
Be = CD
And ∵ △ ABC is an equilateral triangle
ν AB = AC, ∠ ACD = 60 ° i.e. ∠ Abe = ∠ ACD
In △ Abe and △ ACD
BE=CD,∠ABE=∠ACD,AB=AC
∴△ABE≌△ACD（SAS）
∴AE=AD
As shown in the figure, in the right angle trapezoid ABCD, ab ∥ DC, ∠ ABC = 90 °, ab = 2dc, diagonal AC ⊥ BD, vertical foot F, EF ∥ AB, crossing ad at point E, CF = 4cm (1) The results show that the quadrilateral AFE is isosceles trapezoid; (2) Find the length of AE
(1) It is proved that the crossing point D is DM ⊥ ab,
∵DC∥AB，∠CBA=90°，
The quadrilateral BCDM is rectangular
∴DC=MB．
∵AB=2DC，
∴AM=MB=DC．
∵DM⊥AB，
∴AD=BD．
∴∠DAB=∠DBA．
∵ EF ∫ AB, AE and BF intersect at point D, that is, AE and FB are not parallel,
The quadrilateral affe is isosceles trapezoid
（2）∵DC∥AB，
∴△DCF∽△BAF．
∴CD
AB=CF
AF=1
2．
∵CF=4cm，
∴AF=8cm．
∵AC⊥BD，∠ABC=90°，
In △ ABF and △ BCF,
∵∠ABC=∠BFC=90°，
∴∠FAB+∠ABF=90°，
∵∠FBC+∠ABF=90°，
∴∠FAB=∠FBC，
｛ Abf ∽ BCF (AA), namely BF
CF=AF
BF，
∴BF2=CF•AF．
∴BF=4
2cm．
∴AE=BF=4
2cm．
As shown in the figure, AB is the diameter of ⊙ o, and CD is the chord. The vertical lines of CD are drawn through two points a and B respectively, and the vertical feet are e and F. verification: EC = DF
It is proved that O is om ⊥ CD is at point M,
∵OM⊥CD，
∴CM=DM，
∵AE⊥EF，OM⊥EF，BF⊥EF，
∴AE∥OM∥BF，
∵ AB is the diameter of ⊙ o,
∴OA=OB，
/ / OM is the median line of trapezoidal aefb,
∴EM=FM
/ / em-cm = fm-dm, i.e., EC = DF
CD is the chord of ⊙ o, e and F are on the diameter AB, EC is vertical CD, DF is vertical CD, AE = BF is proved
Make om ⊥ CD through point O, cross CD to m, so m is the midpoint of CD. Because EC ⊥ CD, DF ⊥ CD, EC ⊥ om ∥ DF
To sum up, M is the midpoint of CD, EC ‖ om ‖ DF, and EO = of. Because OA = ob, AE = oa-eo, BF = ob-of, AE = BF = BF
In the center O, AB is the diameter of center O, CD is the chord, point E, f are on AB, EC ⊥ CD, FD ⊥ CD verification: AE = BF
Make om ⊥ CD through point O, cross CD to m, so m is the midpoint of CD. Because EC ⊥ CD, DF ⊥ CD, EC ⊥ om ∥ DF
To sum up, M is the midpoint of CD, EC ‖ om ‖ DF, and EO = of. Because OA = ob, AE = oa-eo, BF = ob-of, AE = BF = BF
AB is the diameter of ⊙ o, CD is the chord, AE ⊥ CD is e, BF ⊥ CF is f, EC = DF
Extend AE intersection o to G
Connect OE of og OC od GB
It can be seen that the parallel line segments of EG = BF parallel lines are equal
OG=OB
Angle OGE + OGB = 90
OBD+OBG=90
OBG=OGB
So oBf = OGE
Therefore, the OGE of triangle is equal to oBf
OE=OF
Angle OEF = ofe
CEO = DFO
OCD = ODC OC = OD
Therefore, the triangle OCE is all equal to ODF
Therefore, EC = DF
RELATED INFORMATIONS
- 1. As shown in the figure, AB is the diameter of ⊙ o, CD is the chord, CE ⊥ CD intersects AB with E, DF ⊥ CD intersects AB with F, proving that AE = BF
- 2. As shown in the figure, in the circle O, the chord AB intersects OC, OD at m, n respectively. If AMC = Bnd, it is proved that am = BN
- 3. It is known that in the circle O, AB is the diameter of circle O, CD is a chord, and CD is perpendicular to point ab. connect BC, ad= PA.PB
- 4. As shown in the figure, in the parallelogram ABCD, the bisector of angle ab. C intersects with AD at point P. it is proved that PD + CD = BC
- 5. It is known that, as shown in the figure, the chords AB and CD of circle O intersect at p. it is proved that PA * Pb = PC * PD
- 6. As shown in the figure, the side length of square ABCD is 1cm, e and F are the midpoint of BC and CD respectively. Connecting BF and De, the shadow area in the figure is () cm2 A. 4 Five B. 2 Three C. 5 Six D. 3 Four
- 7. The following is a triangle. Take each vertex of the triangle as the center of the circle and draw an arc with a radius of 2 cm to calculate the shadow area? The shadow is the three corners
- 8. In order to know that the sum of the inner angles of the hexagon ABCDEF is equal, please judge the size of AB plus BC and de plus EF
- 9. As shown in the figure, in the hexagon ABCDEF, AF ∥ CD, ab ∥ ed, ∠ a = 140 °, B = 100 °, e = 90 °. Calculate the degrees of ∠ C, ∠ D and ∠ F
- 10. Root 15 + root 10 - root 5 / root 6 - root 2 + 2 denominator rationalization Is there any simple way to really trouble you
- 11. 1. The radius of circle O is 6cm, the chord AB = 10cm, the chord CD = 8cm, and ab is perpendicular to CD and P
- 12. Given that AB and CD are two parallel chords of circle O, and ab = 48, CD = 40, the distance between the two parallel chords is 22, find the radius of circle o
- 13. As shown in the figure, it is known that in ⊙ o, the chord ab ⊥ CD is e, AE = 2, EB = 8, and the degree of ⊙ CAD is 120 °, then the radius of ⊙ o is______ .
- 14. It is known that in the circle O, if the chord AB / / CD, ab = 2 roots, 11 cm, CD = 4 roots, 5cm, and the distance between the chord AB and CD is 1cm, then the radius of circle O is?
- 15. The radius of circle OA = 10cm, chord AB = 12cm, P is the moving point on AB, find the shortest distance from point P to center O of circle o
- 16. It is known that in circle O, the length of chord AB is three times the root of radius OA, and point C is the midpoint of arc ab. what figure is the quadrilateral oacb and why?
- 17. As shown in the figure, AB is the chord of OD, and the radius OC and OD intersect AB at points E and f respectively, and AE = BF. Verify that OE = of
- 18. As shown in the figure, given Ao ⊥ OC, ob ⊥ OD, ∠ AOB = 142 °, calculate the degree of ∠ cod
- 19. In circle O, given the diameter AB = 2, chord AC length 1 and chord ad length root 2, then the degree of angle DAC is?
- 20. As shown in the figure AB is the diameter of circle O, AC, AD are chords, and ab bisection angle CAD