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If AB is 10, Pb is 4 and OP is 5, then find the radius of the circle
The vertical foot of the vertical line passing through point o as AB is C
Then point C bisects AB AC = BC = 5 so that PC = 1
OC²=OP²-PC²=24
The radius of connection ob is ob 2 = OC 2 + BC 2 = 24 + 25 = 49
So the radius OB=7
If the semicircle of the circle is 5cm and the distance from the center of the circle to the chord AB is 4cm, then AB = how many cm
Dear landlord
Connect ob
In RT △ ODB, OD = 4cm, OB = 5cm
In △ OBD, the Pythagorean theorem is used
Do the rest yourself
I wish you every step of the way
I look forward to your adoption. Thank you
As shown in the figure, in the parallelogram ABCD, e, G, F, h are the points on the four sides respectively, and AE = CF, BG = DH
Proof: connect he, eg, FG, HF
Because the quadrilateral ABCD is a parallelogram
So angle a = angle C
Angle B = angle D
AB=DC
AD=BC
Because ad = ah + DH
BC=BG+CG
So ah + DH = BG + CG
Because BG = DH
So ah = CG
Because AE = CF
So triangle AEH and triangle CGF are congruent (SAS)
So he = FG
Similarly, HF = eg
So the quadrilateral efgh is a parallelogram
As shown in the figure, in the parallelogram ABCD, e, G, F, h are the points on the four sides respectively, and AE = CF, BG = DH
As shown in the figure, in the parallelogram ABCD, e, G, F and H are the points on the four sides respectively, and AE = CF, BH = dg
Connect eh, HF, GF, GE
∵BH=DG,AB=CD
∴AB-BH=CE-DG
Ah = CE
With ? a = ∠ C, AE = CF
∴ΔAEH≌ΔCFE
∴EH=FG
Similarly, HF = eg
The quadrilateral ehfg is a parallelogram
The EF and GH are equally divided
Given that the diameter AB intersects the chord EF with the point P, the angle ape = 45 degrees, and the square of PE + the square of PF = 10, calculate the length of ab To have an explanation, it's a circle. The center of the circle is o
Make om vertical EF, cross EF to M,
EM=MF,
PE^2+PF^2=10,
（EM+PM)^2+(MF-PM)^2=10,
EM^2+2EM*PM+PM^2+MF^2-2MF*PM+PM^2=10,
2EM^2+2PM^2=10,
EM^2+PM^2=5,
AB, CD are the two chords of circle O with radius 5, ab = 8, CD = 6, Mn is the diameter ab ⊥ Mn at point E, CD ⊥ Mn at point F, P as any point on EF Find the shortest distance of PA + PC, AB, Cd on the opposite side of circle o
When the intersection point of BC and EF is p, PA + PC is the shortest
Linking OA, OC, from Pythagorean theorem
OE=3, OF=4
∴EF=7
∵AB‖CD
∴BE/CF=EP/PF
4/3=EP/PF
EP+PF=7
∴EP=4,PF=3
∴BP=4√2, PC=3√2
The shortest distance of PA + PC = BC = 7 √ 2
In △ ABC, ab = 6, AC = 8, BC = 10, P is a moving point on edge BC, PE ⊥ AB is at e, PF ⊥ AC is at F, M is the midpoint of EF, then the minimum value of AM is______ .
∵ a quadrilateral afpe is a rectangle
∴AM=1
When 2AP, AP ⊥ BC, AP is the shortest, and am is also the shortest
When AP ⊥ BC, ⊥ ABP ∷ cab
∴AP：AC=AB：BC
∴AP：8=6：10
The shortest AP, AP = 4.8
When am is the shortest, am = AP △ 2 = 2.4
As shown in the figure, AB and CD are the two chords of ⊙ o with radius of 5, ab = 8, CD = 6, Mn is the diameter, ab ⊥ Mn is at point E, CD ⊥ Mn is at point F, P is on EF
When the intersection point of BC and EF is p, PA + PC is the shortest
Linking OA, OC, from Pythagorean theorem
OE=3,OF=4
∴EF=7
∵AB‖CD
∴BE/CF=EP/PF
4/3=EP/PF
EP+PF=7
∴EP=4,PF=3
∴BP=4√2,PC=3√2
The shortest distance of PA + PC = BC = 7 √ 2
As shown in the figure: (1) Given ab ∥ CD, EF ∥ Mn, ﹤ 1 = 115 ° and find the degrees of ﹤ 2 and ﹤ 4; (2) There is a law implied in this question. Please sum it up according to the result of (1) and try to express it in words; (3) Use the conclusion of (2) to answer: if two sides of two angles are parallel and one angle is twice as much as the other angle, find the size of these two angles
（1）∵AB∥CD，∠1=115°，
∴∠2=∠1=115°，
∵EF∥MN，
∴∠4=180°-∠2=180°-115°=65°；
(2) If two sides of an angle are parallel to each other, then the two angles are equal or complementary;
(3) According to (2), if one of the angles is x, then the other is 2x,
Then x + 2x = 180 °,
X = 60 ° is obtained,
Therefore, the size of these two angles is 60 ° and 120 ° respectively
As shown in the figure, in △ ABC, de ∥ BC, EF ∥ DC, it is proved that ad 2 = ab × AF
∵EF//DC
∴AF/AD=AE/EC
∵DE//BC
∴AD/AB=AE/AC
∴AF/AD=AD/AB
∴AD²=AB×AF
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