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The quadrilateral eofp is a square
∵OE⊥AB,OF⊥CD,AB⊥CD
The shape is rectangular
∵ arc AB= arc BD
∴AB=CD
∴OF=OB
The quadrilateral is square

1. If the bisector ce of the angle OCD intersects the circle E and connects OE, then E is the midpoint of the arc ADB AB is the diameter of circle O, CD is the chord, and CD is vertical AB, and the perpendicular foot is h, 1. The bisector ce of angle OCD intersects E and connects OE. Is e the midpoint of arc ADB 2. If the radius is 1, CD = radical 3 Find the distance from O to chord AC

One
It's the midpoint
Extend CO to intersect o to f
Arc de = arc ef
Arc AD= arc AC= arc FB
Therefore, arc AE = arc EB
That is, e is the midpoint of the arc ADB
Two
The second question is that the length from O to AC is half of BC
In the use of Pythagorean theorem can get Oh = 1 / 2, so BH = 5-1 / 2, and then use Pythagorean theorem to calculate BC length
The second question:
Through H point, do HP vertical AC to P
∵ ob over center CD ⊥ ab
ν ch = HD = 1CD = 2 / 3
In △ och, the angle och = 90 ° ch = the root of 2 / 3 OC = 1
ν Oh = how much is calculated according to Pythagorean theorem
∵OH=. AO=1
ν ah = how much is calculated by yourself
∵CH×AH=AC×HP
Next, I'll do it myself

As shown in the figure, in ⊙ o, the chord AC ⊥ BD, OE ⊥ AB, perpendicular foot is e, verification: OE = 1 2CD．

It is proved that: connect AO and extend the intersection circle to m point, connect MB, MC,

In the circle O, AB is the diameter, the chord CD intersects AB at e, and CE = OE, please guess the relationship between arc BD and arc AC

Connecting OC and OD
∠BOD=∠DEO+∠D
∠DEO=∠C+∠COE
OC=OD
∠D=∠C
CE=OE
∠C=∠COE
∠BOD=∠C+∠COE+∠D=3∠COE
Arc BD = 3 arc AC
☆⌒_ I hope I can help you~

AB is the diameter of circle O, CD is the chord, and CD is perpendicular to AB, the perpendicular foot is h, the bisector ce of angle OCD intersects with E and connects OE. Is e the midpoint of arc ADB?

OE and OC are radii, so the angle OEC = angle OCE,
Because OE is the bisector of the angle OCD, the angle OCE = the angle ECD
So angle OEC = angle ECD, so OE / / CD
Because CD is vertical to AB, OE is vertical to ab
Because AB is the diameter, O is the center of the circle, and OE is perpendicular to AB, e is the midpoint of the arc ADB

As shown in the figure, arc AB is known to be equal to arc CD, OE ⊥ AB, of ⊥ CD, if ⊥ OEF = 25 °, find ≁ EOF step

What about the picture

As shown in the figure, the straight line AB and CD intersect at point O, OE is perpendicular to AB, of is perpendicular to CD, angle AOC = 1 / 4 angle EOF, calculate the degree of angle AOC

It is known that there are: ∠ 1 + ∠ 2 = 90 ° and ∠ 2 = (1 / 4) (∠ 1 + ∠ 2 + ∠ 3)
Therefore: ∠ 1 + ∠ 3 = 180 ° - 2 ∠ 2, and 3 ∠ 2 = ∠ 1 + ∠ 3
So: 3 ∠ 2 = 180 ° - 2 ∠ 2
So: ∠ 2 = 36 degrees
Namely: ∠ AOC = 36 °

As shown in the figure, the straight line AB and CD intersect at point O, OE is perpendicular to AB, of is perpendicular to CD, if angle AOC = quarter angle EOF, calculate the degree of angle AOC

∵OE⊥AB,OF⊥CD
∴∠EOB=∠FOD=90°
A kind of

As shown in the figure, angle AOC: angle cod: angle BOD = 2:3:4, OE, of bisect angle AOC and angle BOD, og bisect angle EOF, and calculate the degree of angle GOF

∵∠AOC:∠COD:∠BOD=2:3:4
∴∠AOC=2k∠COD=3k∠BOD=4k
And ∵ OE and of are equally divided into ∵ AOC and ∵ BOD respectively
∴∠EOF=6k
∵ og bisection ∵ EOF
﹤ GOF = 3K = one third ∠ AOB
If AOB = 180 degrees, GOF = 60 degrees

As shown in the figure, the angle AOB is a flat angle, the angle cod = 60 °, the OE bisection angle AOC, of bisection angle BOD, calculate the degree of the angle EOF

A flat angle minus 60 degrees = 120 degrees
That is, angle AOC + angle BOD = 120
Angle EOF = 1 / 2 (angle AOC + angle BOD) + 60 = 60 + 60 = 120 degrees