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In p-abcd of tetrahedral pyramid, ∠ ABC = ∠ ACD = 90 *, ∠ BAC = ∠ CAD = 60, PA ⊥ ABCD, PA = 2Ab, E.F. are PD, PC midpoint: CE is parallel to plane PA
If your face is parallel to PAB, then make an auxiliary line and make an H point at the middle point of ad.. Connection point EFH
It's up to you to think about it. It's the same as if you've got all your meals and you've got someone to feed you. My friend, think about it,
The quadrilateral ABCD is a rhomboid ∠ ABCD = 30bd = 8piu. Find the degree of ∠ DAB, ∠ ABC, and the length of AB, AC
Since it is diamond and BD = 8, then AC and ab are also 8, but your angle ABCD is god horse situation
The circumference of diamond ABCD is 40cm, and ∠ ABC = 120 °. (1) the degree of abd and DAB. (2) calculate the area of diamond ABCD (3) Find the length of two diagonal lines AC and BD. BD bisection ∠ ABC is known
The circumference of diamond ABCD is 40 cm, ∠ ABC = 120 °
(1) The degree of abd and DAB
∵∠ABC=120°.
ADC = 120 ° (the opposite angles of diamond are equal)
And ? ABC + bad = 180 ° (complementary neighboring angles of diamond)
∴∠DAB=180°﹣∠ABC=60°
(2) Find the area of diamond ABCD
Pass through point D as de ⊥ AB at point E
∵∠DAB=60°
∴∠ADE=30°
The circumference of diamond ABCD is 40 cm
∴AD=40÷4=10﹙㎝﹚
∴AE=10÷2=5﹙㎝﹚
∴DE=√﹙10²-5²﹚=5√3﹙㎝﹚
The area of diamond ABCD: 10 × 5 √ 3 = 50 √ 3 (cm2)
3) Find the length of two diagonals AC and BD
∵ in diamond ABCD,
∴∠ABC=∠ADC=120°,∠DAB=∠DCB=60°
∴BD=10㎝,AC=2×5√3=10√3㎝
As shown in the figure, the radius of circle O is OA = 5cm, AB is a chord, C is a point on AB, and OC is perpendicular to OA, OC = BC. Find the degree of angle A and the length of ab
Connecting ob, the triangle OAB and CBO are isosceles triangle
Angle Bao = ABO, ABO = cob
Bao = ABO angle
In triangle OAB: angle OAB + ABO + BOC + 90 degrees = 180 degrees
Results: OAB of 3 * angle is 90 degrees
So the angle OAB is 30 degrees
AB = 5 * radical 3
As shown in the figure, the chord AC = chord CB, CD is perpendicular to OA in D, CE is perpendicular to ob in e. the process of proving CD = CE should be detailed
Connect OC
⌒ ⌒
AC＝ BC
∠COD＝∠COE
∠ODC＝∠OEC＝90°
OC＝OC
△COD≌△COE
So CD = CE
As shown in the figure: ⊙ o chord CD is perpendicular to diameter AB, e is the midpoint of arc BC, AE intersects CD and CB at g and f respectively, then: () why? A、 AB=CD B、 AF=BF C、 AG=CG D、 CG=CF
Choice: D
prove:
Connect be, let AB and CD intersect M
Because AB is the diameter, ab ⊥ CD
Therefore, e = AMG = 90 degrees
Therefore, a + AGM = CBE + BFE = 90 degrees
Because e is the midpoint of arc BC
So arc be = arc CE
Therefore, a = CBE
Therefore, AGM = BFE
Because ∠ CGF = AGM, ∠ BFE = ∠ CFG
So ∠ CGF = CFG
So CG = CF
In the circle O, the chords AB and CD intersect at point e. it is proved that AE * EB = CE * ed
Because: angle ade = arc AC, angle CBE = arc AC
So: angle ade = angle CBE, and angle AED = angle bec
So: Triangle ade is similar to triangle BCE
That is: AE / CE = de / be
So: AE * EB = CE * De
AB, CD are two mutually perpendicular chords in circle O, with the center angle AOC = 130 ° ad; the extension line of CB intersects at P, and the angle P is calculated
∠ADC=∠ABC=∠AOC/2=65
∠CDP=∠ABP=180-65=115
∠P=360-∠CDP-∠ABP-90=360-115*2-90=40
As shown in the figure, the two chords AB and CD of circle O intersect at point E, EF ∥ CB, EF intersects the extension line of ad at point F, FG tangent circle O at point G, EF = 2, then the length of FG is () A. 1 Two B. 1 Three C. 1 D. 2
∵EF∥CB，∴∠DEF=∠C．
∵ in circle O, ᙽ a, ᙽ C are the same as the opposite arc BD, a = ∠ C
So ∠ def = ∠ a,
DFE = ∵ EFA, DFE ∵ EFA, FD is obtained
EF＝EF
FA
∴EF2=FD•FA，
∵ FG tangent o at point G, ᙽ fg2 = FD · FA, then EF = FG
The length of FG is 2
Therefore, D
AB is the diameter of the circle, chord CD is perpendicular to point E, point P is on the circle, angle CBP = angle BCD, CB is parallel to PD, if BC = 3, the sine value of angle BPD is equal to 3:5 Find the diameter of a circle
∠BPD=∠BCD
∵sin∠BPD=3/5
∴BE/BC=sin∠BCD=3/5
∵BC=3
∴BE=9/5 CE=12/5
Connecting AC
△ACB∽△CED
∴BE/CB=CB/AB
AB=CB*CB/BE=3*3/(9/5)=5
The diameter of the circle is 5
RELATED INFORMATIONS
- 1. AB is the diameter of circle O, the chord CD is perpendicular to e, P is on the circle, ∠ CBP = ∠ C, find CB / / PD
- 2. AB is the diameter of circle O, AC is the chord, OD ⊥ AC is at D, and the tangent AP of circle O is made through a. the extension line of AP and OD intersects P, connecting PC and BC (1) (2) prove that PC is tangent of circle o
- 3. 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
- 4. 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?
- 5. 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
- 6. 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
- 7. If the length of a string is equal to the radius, then the radians of the central angle of the chord are? Why?
- 8. 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?
- 9. The third degree of the sixth root is not equal to 5:10 of the third part of the root sign
- 10. Given that circle O1 and circle O2 are inscribed at point a, the chord ab of circle O1 intersects with circle O2 at point C, the ratio of radius of circle O1 to circle O2 is 2:3, ab = 12 emergency
- 11. 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
- 12. 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
- 13. 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
- 14. 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?
- 15. 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
- 16. As shown in Fig. 3, in the triangle ABC, ad bisects ∠ BAC, passing through B is vertical ad, perpendicular to foot is point E, crossing e is EF / / AC, and ab is intersected with F, AF and BF
- 17. As shown in the figure, △ ABC, ab = AC, ⊙ o with ab as diameter intersects BC at point P, PD ⊥ AC at point D (1) It is proved that PD is tangent of ⊙ o; (2) If ∠ cab = 120 ° and ab = 2, calculate the value of BC
- 18. 0
- 19. As shown in Fig. 1, ⊙ o is the circumscribed circle of ⊙ ABC, and ∠ B = ∠ CAD
- 20. As shown in the figure, given the triangle ABC, the angle a = 90 degrees, take AB as the diameter to make the semicircle intersect BC at point D, and cross point D as the tangent of circle O to intersect AC at point P. it is proved that PA = PC