As shown in Fig

As shown in Fig

Let the tangent point of tangent CD and circle be e
Then AC = CE, BD = De, PA = Pb
ν PD + CD = perimeter
=PC+CE+DE+PD
=PC+AC+ BD+PD
=PA+PB
=PA+PA
=12

As shown in the figure, PA and Pb cut circle O to a and B respectively, and intersect with tangent lines of circle O at C and D respectively. It is known that the circumference of △ PCD is equal to 10cm, then PA=______  cm．

As shown in the figure, set the tangent point of DC and ⊙ O as E;
∵ PA and Pb are tangent lines of ⊙ o respectively, and the tangent points are a and B;
∴PA=PB；
Similarly, it can be concluded that de = Da, CE = CB;
Then the circumference of △ PCD = PD + de + CE + PC = PD + Da + PC + CB = PA + Pb = 10 (CM);
∴PA=PB=5cm，
So the answer is: 5

As shown in the figure, PA, Pb, De, respectively tangent o to a, B, C, the radius of circle O is 6cm, Po = 10, then the circumference of triangle PDE is

PA²=PO²-OA²=100-36=64
PA=8
So Pb = PA = 8 (PA, Pb are tangent lines through P)
DC = Da, EC = EB
So the circumference of the triangle PDE = PD + de + PE = PD + DC + CE + PE = PD + Da + EB + PE = PA + Pb = 16 cm

As shown in the figure, ⊙ o is the circumscribed circle of RT △ ABC, ∠ ABC = 90 °, point P is a point outside the circle, PA cuts ⊙ o at point a, and PA = Pb (1) It is proved that Pb is tangent of ⊙ o; (2) Known pa= 3, BC = 1, find the radius of ⊙ o

(1) The ? Pao, ? PbO, ? PbO, ? PbO, ? PbO, ? PbO, ? PbO, ? PbO, ? PbO, ? PbO

As shown in the figure, circle O is the circumscribed circle of RT △ ABC, ∠ ABC = 90 degrees, point P is a point outside the circle, PA tangent circle O to point a, and PA = Pb (1) prove: Pb is the tangent of circle O (2) known PA = radical 3, BC = 1 [1] find the distance between P and o [2] find the length of arc ab

The center of the circle is o
Link OP, ob
Because it's the radius of the circle, OA = ob
It is known that PA = Pb and share the edge Op
The result shows that the triangle OPA is equal to triangle OPB, and the angle OBP is 90 degrees, and Pb is the tangent line of circle o

Let | e be a moving point in the graph | 2, where | e is a moving point The straight line m is perpendicular to o, Ao = Bo. 1) establish a suitable coordinate system and solve the equation of curve E. 2) let d be a point on the line M

That is, the ellipse with focal length C = 1 / 2|ab| = 1, over (- 1, Radix 2 / 2)
The coordinate system is established with O origin, and M is the Y axis
Let e: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1
a^2-b^2=1/2 (1)
1/a^2+1/2b^2=1 (2)
A = radical 2, B = 1
E:x^2/2+y^2=1
That's where d comes from

As shown in the figure, OA ⊥ ob, OC ⊥ OD, ∠ AOD = 4 ∠ BOC, and calculate the degree of ∠ BOC

∵ OA ⊥ OC, ob ⊥ OD,  BOD = 90 °, AOC = 90 °, BOD + AOC = 180 °, i.e.,  cod + ∠ BOD + ∠ AOD + ∠ BOC = 180 °, and ∵ AOD + BOC = 180 °, and ? AOD = 4 ﹤ BOC, ? BOC = 45 °. No, looking for me is my reward!

OA is perpendicular to ob, OC is perpendicular to OD, if ∠ AOD = 140 °, then ∠ BOC = () degree?

OA is perpendicular to ob, OC is perpendicular to OD, if ∠ AOD = 140 °, then ∠ BOC = (40) degrees
∠BOC＝360°－∠AOB－∠AOD－∠DOC＝360°－140°－90°－90°＝40°

As shown in the figure, OA ⊥ ob, OC ⊥ OD=______ .

∵OA⊥OB，OC⊥OD，
∴∠AOB=∠COD=90°；
And ? AOD + ∠ AOB + ∠ BOC + ∠ cod = 360 °, AOD = 144 °,
∴∠BOC=36°；
So the answer is: 36 degrees

Six rays OA, ob, OC, OD, OE, of are drawn from point O, and ∠ AOB = 90 ° of bisection ∠ BOC, OE bisection ∠ AOD, if ∠ EOF = 170 °, calculate the degree of ∠ cod

70°
Let ∠ cod = x ∠ EOD + ∠ cod + ∠ COF = 170 = ∠ EOF
Since of bisects ∠ BOC, ∠ COF = ∠ BOF
Because OE bisects ∠ AOD, so ∠ EOD = ∠ EOA
So ∠ BOF + ∠ EOA = 170-x
Because ∠ BOF + ∠ EOA + ∠ EOF + ∠ AOB = 360 ∠ AOB = 90 ∠ EOF = 170
So 170-x + 170 + 90 = 360, and finally x = ∠ cod = 70