AB, CD are the two chords in the circle, AB is perpendicular to e, AE = 2, EB = 6, ed = 4, then the diameter of the circle is The distance between the tangent points of isosceles triangle with waist 10 and base length 6 is 0

AB, CD are the two chords in the circle, AB is perpendicular to e, AE = 2, EB = 6, ed = 4, then the diameter of the circle is The distance between the tangent points of isosceles triangle with waist 10 and base length 6 is 0

1. AE = 2, EB = 6. According to the vertical diameter theorem, AB bisects CD, and AE + EB = AB, so AB = 2 + 6 = 8
2. Between the tangent points of the two waists, it refers to the median line of the isosceles triangle, so d = 1 / 2, bottom length = 0.5 * 6 = 3

As shown in the figure, we know: in ⊙ o, the diameter ab ⊥ CD, e is the perpendicular foot, AE = 4, CE = 6, then the radius of ⊙ o is______ .

Connect OC, ab ⊥ CD, so △ OCE is a right triangle,
Let OC = x, then x is obtained from Pythagorean theorem=
CE2+OE2＝
62+(x−4)2，
The results show that, 8x = 52,
The solution is x = 6.5,
The radius of ⊙ o is 6.5
So the answer is: 6.5

AB is the diameter of ⊙ o, BC, CD, DA are the chords of circle ⊙ and BC = CD = da. Find the degree of angle BOD

Because BC = Da, ABCD is isosceles trapezoid
∠ODB=∠OBD=∠CDB=∠CBD
So △ OBD ≌ △ CBD
So △ oad is an equilateral triangle
Angle BOD = 120 degrees

As shown in the figure, AB is the diameter, the chord CD ⊥ AB is at point E, ≁ CDB = 30 ° and the radius of ⊙ o is The length of the string CD is 3 cm______ cm．

∵∠CDB=30°，
∴∠COB=30°×2=60°．
The radius of O is
3cm，
∴CE=
3sin60°=
3 x
Three
2=3
2，
∴CD=3
2×2=3（cm）．

As shown in the figure, in the two concentric circles with point o as the center, the chord ab of the large circle is the tangent line of the small circle, and the point P is the tangent point

Proof: connect OP as shown in the figure,
∵ the chord ab of the big circle is the tangent line of the small circle, and the point P is the tangent point,
∴OP⊥AB，
∵ OP over O,
∴AP=BP．

Take the chord ab of ⊙ o as the edge and make a square ABCD outside the circle (1) As shown in Figure L, OC = OD; (2) As shown in Figure 2, make DM cut ⊙ o to m through D, if AB = 2, DM = 2 2. Find the radius of ⊙ o

(1) Connect OA, ob, ∵ OA = ob,  OAB = ∵ oba, ∵ ABCD is a square,  DAB = ∠ ABC = 90 ° and ﹤ oad = ∠ OBC. In △ OAD and △ OBC, OA = ob ∵ oad = ob CAD = BC,  OAD ≌ △ OBC (SAS),

As shown in the figure, AB is known to be the diameter of ⊙ o, and the chord CD ⊥ AB, AC = 2 2, BC = 1, then the value of sin ∠ abd is______ .

∵ AB is the diameter of ⊙ o,
∴∠ACB=90°，AB=
12+(2
2)2=3．
∴sin∠ABD=sin∠ABC=AC
AB=2
Two
3．

As we know in the figure that AB is the diameter of circle O, the chord CD is perpendicular to ab AC = 2 times the root 2 BC = 1, then the value of cos ∠ abd is

I don't know if your picture is the same
∵ AB is the diameter
AB⊥CD
ν de = EC (vertical diameter theorem)
ν BD = BC (three wires in one)
∴∠ABD=∠ABC
AC=2√2 BC=1
The Pythagorean theorem AB = 3
∴cos∠ABD=cos∠ABC=BC/AB=1/3

As shown in the figure, in circle O, the chord AB and CD intersect at P, 1, if AB, CD and op form equal angles, verification: ab = CD 2, if AB = CD, verify: AC = BD; PA = PD

Proof: 1
O is used as OE ⊥ AB at point E and o as of ⊥ CD at point F
In the right triangle ope and the right triangle OPF
∵ AB, CD is equal to Op
∴∠OPE=∠OPF
And OP is the public side
≌ right triangle op ≌ right triangle OPF (angle, angle, edge)
Therefore, OE = of ① PE = PF ②
In right triangle AOE and right triangle COF
OE = of is obtained from ①
OA, OC are the radius of the circle
So OA = OC
≌ right triangle AOE ≌ right triangle COF (hypotenuse, right angle side)
Thus AE = CF 3
PE + AE = pf + CF was obtained from ② + ③
AP = CP 4
According to the intersecting chord theorem of circles, AP * Pb = Cp * PD ⑤ is obtained
From (4) to (5), Pb = PD (6)
④ AP + Pb = CP + PD
AB = CD
2．
Connect AC, BC, BD, ad
In triangle BAC and triangle CAD
∵AB=CD
ν arc AB = arc CD
Thus ∠ ACB = ∠ CAD (equal arc and equal circle angle)
Arc BC= arc CD- arc BD, arc AD= arc AB- arc BD
ν arc BC = arc ad
Thus, ﹤ ACD= ﹤ BAC (the circular angles of equal arcs are equal)
And AC is the public side
≌ triangle BAC ≌ triangle CAD (angle, edge, angle)
Thus AC = BD
In triangle PAC and triangle PBD
∠CAP=∠CDB,∠ACP=∠DBP
It has been proved that AC = BD
≌ triangle PAC ≌ triangle PBD (angle, edge, angle)
So PA = PD

In a circle of radius 1, the length is equal to The chord of 2 is at the center of the circle______ Degree

As shown in the figure, in ⊙ o, ab=
2，OA=OB=1，
∴AB2=OA2+OB2，
Ψ Δ AOB is a right triangle, and ∠ AOB = 90 °,
That is, the length is equal to
The chord of 2 is at 90 ° to the center of the circle
So the answer is: 90