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?

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?

Do the midpoint e of CD, OE, vertical CD, connect OC, OD
OE=OP/2=2
The square of ED + the square of OE = the square of OD
Therefore, ed = 2 root sign 15
CD = 4 root numbers 15

As shown in the figure, AB and CD are two strings, and arc ad = arc BC, BP = DP

∵ arc ad = arc BC
ν arc AB = arc ad - arc BD = arc BC - arc BD = arc CD
BP = DP
ν arc BP = arc DP
ν arc AP = arc BP - arc AB = arc DP - arc CD = arc CP
Get the certificate

If point P is the point on the chord AB, connect OP, pass P as PC ⊥ OP, and PC intersect ⊙ o at point C. If AP = 4, Pb = 2, then the length of PC is () A. Two B. 2 C. 2 Two D. 3

Extend CP to ⊙ o at point D,
∵PC⊥OP，
∴PC=PD，
∵PC•PD=PB•PA，
∴PC2=PB•PA，
∵AP=4，PB=2，
∴PC2=8，
The length of the PC is: 2
2．
Therefore, C

As shown in the figure, P is the point on the chord AB, CP ⊥ OP intersects ⊙ o at point C, ab = 8, AP PB=1 3. Find the length of PC

As shown in the figure, extend CP to d
∵CP⊥OP，
∴CP=DP．
∵AB=8，AP
PB=1
3，
∴AP=1
4AB=2，PB=3
4AB=6．
∵ AB and CD are two intersecting chords of ⊙ o, and the intersection point is p,
∴PC•PD=AP•PB，
∴PC2=2×6，
∴PC=2
3．

As shown in the figure, in circle O, the chords AB and CD intersect vertically at point E In the right triangle ace, ∠ bad + ∠ ACD = 90 ° why

There is a theorem that you must know first: the angle of the center of a circle opposite the same arc is twice the angle of the circle opposite to the same arc. Prove:

As shown in the figure, AB is a chord of ⊙ o, OD ⊥ AB, perpendicular foot is C, intersect ⊙ o at point D and point E on ⊙ o (1) If ∠ AOD = 52 °, calculate the degree of ∠ DEB; (2) If OA = 5, ab = 8, find the size of Tan ∠ AEB

(1) ∵ OD ⊥ AB, ∵ ad = dB,  DEB = 12 ⊥ AOD = 12 × 52 ° = 26 °. (4 points) (2) ∵ OD ⊥ AB, ᙽ arc ad = arc BD = 12 arc AB, ? AC = BC = 12ab = 4,  AOC is a right triangle, ? AEB = ? OA = 5. From the Pythagorean theorem, OC = 52 ⊥ 42 = 3,  Tan ? AEB =

As shown in the figure, AB is a chord of ⊙ o, OD ⊥ AB, perpendicular foot is C, intersect ⊙ o at point D, point E on ⊙ o. ∠ AOD = 52 ° and ⊙ DEB=______ .

⊙ in ⊙ o, OD ⊥ AB,
Qi
AD=
BD，
∵∠AOD=52°，
∴∠DEB=1
2∠AOD=26°．
So the answer is: 26 degrees

As shown in the figure, in circle O, the radius od is perpendicular to chord AB, and the perpendicular foot is C,

Analysis: this problem uses the vertical diameter theorem, the relationship between the circular angle and the central angle, as well as the Pythagorean theorem

As shown in the figure, if there is a chord ab of length 4 in the circle O with radius 2, then the degree of the center angle of the circle to which the chord AB corresponds is? As shown in the figure, if there is a chord ab of length 4 in the circle O with radius 2, then the degree of the center angle of the chord AB is A.60°； B.90°； C.120° ； D.150°.

No, the diameter corresponds to the center angle of the circle, which is 180 °

As shown in the figure, BC is the diameter of ⊙ o, a is a point on the extension line of chord BD, and the tangent de bisects AC to e. it is proved that AC is the tangent of ⊙ o

Proof: connect OD, OE, CD;
∵ tangent de bisects AC to E,
∴∠ODE=90°，
∵ BC is the diameter of ⊙ o,
∴∠BDC=∠ADC=90°，
∵AE=EC，
In RT △ ADC: de = CE = 1
2AC；
∵OE=OE，OD=OC，
∴△ODE≌△OCE，
∴∠ACB=90°，
⊙ AC is the tangent line of ⊙ o