It is known that AC is the diameter of circle 0, PA is vertical AC, connecting OP, xuancb is parallel OP, the straight line Pb intersects the straight line AC at point D, BD = 2PA

D on AC extension line
Connect ob, AP = BP
sinD=1/3
OB=OA=1/3*BFD
Tan ∠ OPA = 2 / 3, sin ∠ OPA = 2, root number 13 / 13

Known: AC is the diameter of circle O, PA is perpendicular to AC, connect OP, chord Pb intersect straight line AC and D, BD = 2PA, find the value of sin ∠ OPA~ The method is as simple as possible Where is the capital of mathematics? I asked several questions, but no one answered them~ Sorry, I copied the serial~ Revise again~ It is known that: AC is the diameter of circle O, PA is perpendicular to AC, connecting OP, chord CB is parallel to OP, straight line Pb intersects straight line AC to D, BD = 2PA (PS: it has been proved in the first two questions that Pb is the tangent line of circle O, Po = 3 / 2PB, no need to prove) Find the value of sin ∠ OPA

If the radius is 1, then Ao = 1, ad = 4, pad is a right triangle. Let PA = x Pb = x BD = 2x, that is, PD = 3x,
Pythagorean theorem PA square + ad square = PD square, get x = root 2, further Po = root 3, your answer will die 3

As shown in the figure, two tangent lines are introduced from a point p outside circle O to circle O. the tangent points are A.B. the diameter AC of the circle is made through point a, and CB is connected to verify CB ∥ Op

∵∠AOB=∠BOC+∠COB,∠BOC=∠COB
∴∠AOB=1/2∠CBO
RT ⊿ AOP, RT ⊿ BOP
∵OP=OP,OA=OB
∴RT⊿AOP≌RT⊿BOP
∴∠AOP=∠BOP
∵∠AOB=∠AOP+∠BOP
∴∠BOP=∠CBO
∴CB‖OP

As shown in the figure, ⊙ o is the circumscribed circle of △ ABC, the chord CD bisects ∠ ACB, ∠ ACB = 120 ° and finds Ca + CB CD value

Connect AD and DB; make be ∥ CD to AC extension line at E. ? CD bisection  ACB, ? ACB = 120 °,  e = ∠ ACD = 60 °, ∠ ECB = 60 °, △ BEC is an equilateral triangle, ? be = EC = CB, ? ADB = 180 ° -  ACB = ∠ ECB = 60 °, ad = BD, ﹤ ADB is an equilateral triangle,  DB = AB, in

Let p be a point above the line CB, CA ⊥ CB, PA ⊥ Pb, and PA = Pb, PM ⊥ BC in M, if CA = 1, PM = 4. Find the length of CB

This question is divided into the following two situations:
① As shown in Fig. 1, PN ⊥ CA is used in N through P,
∵PA⊥PB，
∴∠APB=90°，
∵∠NPM=90°，
∴∠NPA=∠BPM，
In △ PMB and △ PNA,
∠N＝∠BMP
∠NPA＝∠BPM
PA＝PB ，
∴△PMB≌△PNA，
∴PM=PN=4=CM，BM=AN=3，
∴BC=7；
② As shown in Fig. 2, PN ⊥ CA is used in N through P,
∵PA⊥PB，
∴∠APB=90°，
∵∠NPM=90°，
∴∠NPA=∠BPM，
In △ PMB and △ PNA,
∠N＝∠BMP
∠NPA＝∠BPM
PA＝PB ，
∴△PMB≌△PNA，
∴PM=PN=4=CM，BM=AN=5，
BC = 9
Combined with the above CB = 7 or 9

Let p be a point above the line CB, CA ⊥ CB, PA ⊥ Pb, and PA = Pb, PM ⊥ BC in M, if CA = 1, PM = 4. Find the length of CB

As shown in the figure, BC is the diameter of circle O, P is a point on CB extension line, PA tangent circle O to a, if PA = √ 3, Pb = 1, find the radius of circle o

Let the radius be r, then: Po = Pb + Bo = R + 1, Ao = R, so R ^ 2 + 3 = (R + 1) ^ 2, r = 1, so the radius of the circle is 1. I hope it can help you,

Given that AB is the chord of ⊙ o, P is a point on AB, ab = 10, PA = 4, Op = 5, find the radius of ⊙ o

O is used as OE ⊥ AB, and the vertical foot is e, connecting OA,
∵AB=10，PA=4，
∴AE=1
2AB=5，PE=AE-PA=5-4=1，
In RT △ Poe, OE=
OP2−PE2=
52−12=2
6，
In Rt △ AOE, OA=
AE2+OE2=
52+(2
6)2=7．

As shown in the figure, CD is the chord of the center O, AB is the diameter, CD ⊥ AB, and the perpendicular foot is p. it is proved that the square of PC = PA * Pb

CD ⊥ AB PC ⊥ AB angle ACB is right angle AC ⊥ BC triangle ACP and triangle ACB are similar to triangle BCP, angle ACP = angle PBC, so tgacp = tgpbc is proved because tgacp = AP / PC, tgpbc = PC / Pb AP / PC = PC / Pb PC * PC = AP * Pb

As shown in the figure: Mn is the tangent line of ⊙ o, a is the tangent point, passing through point a is AP ⊥ Mn, the chord BC of ⊙ o is at point P, if PA = 2cm, Pb = 5cm, PC = 3cm

Extend AP intersection ⊙ o at point D;
∵PA•PD=PC•PB，
∴2×PD=3×5，
∴PD=7.5cm，
⊙ the diameter of O is ad = PA + PD = 2 + 7.5 = 9.5cm

It is known that AC is the diameter of circle 0, PA is vertical AC, connecting OP, xuancb is parallel OP, the straight line Pb intersects the straight line AC at point D, BD = 2PA