As shown in Figure AC is the diameter of circle O, PA is perpendicular to AC, connecting OP, chord CB / / op, diameter BC intersects straight line AC at D, BD = 2PA, prove that BP is tangent of circle O, OP is related to BC Verify the value of sin angle OPA I 1 and can not insert the picture The diameter BC in the problem is changed to the straight line Pb

1. Connect ob
Because CB ‖ op
So ∠ BCO = ∠ POA
Because ob = OC
So ∠ BCO = ∠ CBO
So ∠ CBO = ∠ POA
Because ∠ CBO = ∠ POB
So ∠ BOP = ∠ POA
In △ POB and △ POA
PO=PO
∠BOP=∠POA
OB=OA
So △ POB ≌ △ Poa
So ∠ PbO = ∠ Pao = 90 °
So BP is tangent to circle o
Two

As shown in the figure, PA and Pb cut ⊙ o at points a and B, and point C is a point above ⊙ o, and ∠ ACB = 65 °, then ∠ P=______ Degree

Connect OA, ob
When PA and Pb cut ⊙ o at points a and B, then ∠ Pao = ∠ PbO = 90 °,
According to the circular angle theorem, ∠ AOB = 2 ∠ C = 130 °,
∵∠P+∠PAO+∠PBO+∠AOB=360°，
∴∠P=180°-∠AOB=50°．

As shown in the figure, PA and Pb are tangent to a and B respectively, and AC is the diameter

Connect ab. cross OP to E
Because PA and Pb are tangent lines of a circle, PA = Pb, and the triangle PAB is an isosceles triangle
It is proved that the triangle Pao is congruent with the triangle PbO. Therefore, according to the theorem of three lines in an isosceles triangle, OP is perpendicular to ab,
AC is the diameter, so AB is perpendicular to BC,
So the angle AEO = angle ABC = 90 degrees,
So BC / / op

As shown in the figure, P is the point on the chord ab of ⊙ o, PA = 6, Pb = 2, and the radius of ⊙ o is 5, then Op=______ .

As shown in the figure, connect OA, make OC ⊥ AB through point O, and C for vertical foot,
∵PA=6，PB=2，
∴AC=4，
∴PC=2，
∵OA=5，
According to Pythagorean theorem, OC = 3,
∴OP=
OC2+PC2=
22+32=
13．
So the answer is:
13．

As shown in the figure, AB is the chord of O, P is on AB, ab = 10, PA = 6, the radius of O is 7, find the length of Op The picture can't be transmitted. There is a circle. The center of the circle is point O. there is a string ab. on the circle O, there is a point P on ab You can get extra points if you get the answer

There are many methods. In order to explain it conveniently, Pythagorean theorem is used here
The intersection point of the vertical line passing through the center of the circle AB is C, and C is also the midpoint of AB, that is, AC = 5, CP = 1,
Using the Pythagorean theorem:
AC=(OA^2-AC^2)^0.5=24^0.5;
OP=(AC^2+CP^2)^0.5=5
So the op length is 5

As shown in the figure, CD is the chord of ⊙ o, AB is the diameter, CD ⊥ AB, and the perpendicular foot is p. it is proved that PC2 = PA · Pb

Proof: connect AC, BD,
∵∠A=∠D，∠C=∠B，
∴△APC∽△DPB．
∴CP
BP=AP
DP，
∴CP•DP=AP•BP．
∵ AB is the diameter, CD ⊥ AB,
∴CP=PD．
∴PC2=PA•PB．

As shown in the figure, introduce two tangents PA, Pb from point P to ⊙ o, tangent points are a, B, BC is the diameter of ⊙ o, AC is the chord, if ∠ P = 60 ° Pb = 2cm, find the length of AC

As shown in the figure: connect ab
∵ PA, Pb are tangent lines,
∴PA=PB．
And ∵ P = 60 °,
∴AB=PB=2cm．
∵ BC is the diameter,
∴∠BAC=90°．
And ∵ CB ⊥ Pb, and ∵ PBA = 60 °,
∴∠ABC=30°．
Then AC = abtan30 ° = 2 ×
Three
3=2
Three
3 (CM), i.e. the length of AC is 2
Three
3cm．

As shown in the figure, BC is the diameter of ⊙ o, P is a point outside ⊙ o, PA and Pb are tangent lines of ⊙ o, and the tangent points are a and B respectively

It is proved that AB is connected to op to F, and Ao is connected
∵ PA, Pb are tangent lines of a circle,
∴PA=PB，
∵OA=OB
The Po is vertically bisected ab
∴∠OFB=90°．
∵ BC is the diameter,
∴∠CAB=90°．
∴∠CAB=∠OFB．
∴AC∥OP．

Point P is a point on circle O, chord AB is vertical bisector OP, and ab = radical 3, point D is any point of arc APB, (not coincident with a and b), de ⊥ AB is at point E Take D as the center of the circle and take the length of de as the radius to make the circle D. pass through a and B respectively to make the tangent lines of circle D. the two tangents intersect at C (1) Find the radius of circle O (2) Find the size of ∠ ACB (3) Let's note that the area of △ ABC is s, if s = the square of 3DE, find AC + BC The first two questions have been done by myself. The radius is 1 and the angle ACB is 60 degrees It's time for the high school entrance examination,

3) With AD, BD, CD,
Because de ⊥ AB is at point E, make a circle d with the length of de as the radius
So AB and D tangent to E,
Then go through a and B to make the tangent of circle D
So the circle D is the inscribed circle of △ ABC,
Because s = △ abd area + △ BCD area + △ ACD area
=(1/2)*AB*DE+(1/2)BC*DE+(1/2)AC*DE
=(1/2)DE(AB+BC+AC)
So (1 / 2) de (AB + BC + AC) = 4 root sign 3DE
So √ 3 + BC + AC = 8 √ 3
So AC + BC = 7 √ 3

As shown in the figure, the vertical bisector of chord AB intersects at point C and intersects chord AB at point D. It is known that ab = 24cm, CD = 8cm. (1) calculate the circle where the fragment is located (do not write the method, keep the drawing trace) (2) calculate the area of the circle where the fragment is located

(1) Methods: connect BC, make the vertical bisector of BC, intersect with the straight line CD at point O, draw a circle with point o as the center and OC as the radius to draw a circle, that is to say, connect OA