As shown in the figure, in the circle O with radius r, the angle AOB is equal to 2a, and OC is perpendicular to ab at point C. find the length of the chord AB and the distance between the chord centers

Angle AOB + angle a + angle B = 180 degree
Because the angle AOB equals 2A
Angle a = angle B
So it can be concluded that
2a+a+a=180°
Angle a = 45 degrees
Angle AOB = 90 degree
ab=r√ 2
Chord center distance OC = R / √ 2

As shown in the figure, ad and BC intersect at O, OA = OC, ∠ a = ∠ C, be = De

It is proved that in △ AOB and △ 1od, ∠ a ＝ 1oa = O1 ∠ AOB ＝ 1od, ≌ AOB ≌ 1od (ASA), ≌ ob = OD, ≌ point O is on the vertical bisector of line BD, ∵ be = De, ≌ point E is on the vertical bisector of line BD and ≌ OE is on the vertical bisector of line BD

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