Because the quadrilateral OABC is an isosceles trapezoid, AB = 4, ∠ COA = 60.
Therefore: OC=AB=4, ∠ OAB = 60, AN=OM, CM=BN.
So: OM= 1/2? OC=2=AN，CM=2√3=BN
Because BC//OA, OA=7
Therefore: MN=OA-OM-AN=3.
Therefore: ON=OM+MN=5.
Therefore: b (5 5,2 √ 3)
(2) If △OCP is an isosceles triangle, because ∠ COA = 60.
Then: △OCP is a regular triangle or P is on the negative semi-axis of the X axis.
① When △OCP is a regular triangle,
Therefore: OP=OC=4.
Therefore: p (4 4,0)
② When P is on the negative semi-axis of the X-axis
And OP=OC=4.
Therefore: p (-4,0)
(3)CPD =∠OAB =∠COA = 60
Therefore: ∠ OPC+∠ DPA = ∠ DPA+∠ ADP = 120.
Therefore: ∠OPC=∠ADP.
Therefore: △OPC∽△ADP
Therefore: OP/AD=OC/PA.
Because 8*BD=5*AB, AB=4.
Therefore: BD=5/2.
Therefore: AD=AB-BD=3/2.
Let OP =x, so: pa = OA-op = 7-X.
Therefore: x/(3/2)=4/(7-x)
So: x= 1 or x=6.
Therefore: p (1, 0) or p (6, 0)