(1)⊿adc in the first quadrant: connect BD and OD, and OD passes through AB to E.

∠AOD=β, that is ∠ AOD = ∠ ABO, ∠ AOD+∠ OAB = ∠ ABO +∠ OAB = 90.

∴∠aeo=90； AD=AO, then: ∠OAB=∠DAB. (The height of the base of an isosceles triangle is also the bisector of the vertex)

And AB=AB, then: ⊿ OAB ≌ δ dab (SAS), ∠ ADB = ∠ AOB = 90.

∴∠ ADB+∠ ADC = 180, and points B, D and C are on the same straight line. -? So we just need to find the analytical formula of straight BD.

OE⊥AB in E, the area relation shows that AB*OE=OA*OB, that is, 5 * OE = 3 * 4, OE = 12/5, OD = 2OE = 24/5.

Let the DF⊥X axis be in F. Yi Zhi: ⊿ ofd ∽⊿ boa, df/ao = OD/BA, df/3 = (24/5)/5 and df = 72/25.

Similarly: OF=96/25, that is, d is (96/25, 72/25).

Starting from the coordinates of B (0 0,4) and d (96/25,72/25), the straight line BD (CD) is: y=(-7/24)x+4.

(2) When ⊿ADC rotates to ⊿AD'C' in the fourth quadrant: c' is (0, -4), connecting AC' and od'.

It is easy to know that ∠ AC 'o = ∠ ABO = β; And ∠AOD'=β, then: ∠ aod' = ∠ AC 'o

Similarly, we can know AC' ⊥ OD', ⊿ AD 'C' ≌ δ AOC' ≌ δ ADC.

Similarly, we can get: d' is (96/25, -72/25). - ? At this time, d' and d are symmetrical about X.

Similarly, the straight line C'D' is: y=(7/24)x-4.

To sum up, the straight line CD is y=(-7/24)x+4 or y=(7/24)x-4.