OA is the radius of circle O, OP is perpendicular to OA, the chord AB intersects op with C, and Pb = PC

OA is the radius of circle O, OP is perpendicular to OA, the chord AB intersects op with C, and Pb = PC

PB=PC,∠PBC=∠PCB.
OB=OA,∠ABO=∠BAO,
∠PCB=∠ACO,
OP vertical OA, ∠ ACO + ∠ Bao = 90 °,
∠PBO=∠PBC+∠ABO=90°,
PB⊥BO,
Pb is the tangent of circle o

As shown in the figure, if the chord BC passes through the midpoint P of the radius OA of the circle O, and Pb

B,8

In the circle O, if the chord AB is equal to the radius OA, then the ratio of the arc length of the inferior arc to the chord length and the chord length to the length of the chord ab

∵ connect ob
Then AB = OA = ob = R (radius)
The OAB is an equilateral triangle
∴∠AOB=60°=π/3
Arc AB = π R / 3
Arc AB: ab = π / 3

In the circle O, the center angle AOB = 60 degrees, C is a point on the arc AB, passing through the point C as cm perpendicular to OA and m, CN perpendicular to ob and n
In the circle O, the center angle AOB is 60 degrees, the radius is 10 (other values can be known and fixed anyway), C is a point on the arc AB, passing through the point C, CM is perpendicular to OA and m, CN is perpendicular to ob and n. It is proved that no matter point C is any point on the arc AB, the length of Mn is fixed

Extend cm circle O to D, extend CN circle O to e, and connect De
From AOB = 60 degrees, CM is perpendicular to OA, CN is perpendicular to ob
It can be seen that MCN = 120 degrees = DCE
So the chord length is fixed
Because cm is perpendicular to OA, CN is perpendicular to ob
So cm = MD, CN = ne
So Mn = 1 / 2DE (median line)