S△ABP=AB×PM/2，S△BPC=BC×PN/2，

∴S△ABP/S△BPC=AB/BC=30/23

(2)AOD =∠AOE

Proof: ∠ DAC = ∠ DAB+∠ BAC, = 60+∠ BAC, ∠ BAE = ∠ EAC+∠ BAC = 60+∠ BAC,

∴∠DAC=∠BAE,AD=AB,AC=AE,∴△DAC？ △BAE(SAS),∴∠ACD=∠AEB，

Let AC be q, ∠ AQE = ∠ OQC, ∠ EAQ =180-(∠ AQE+∠ AEB),

∠cbq = 180 -(∠oqc+∠acd),∴∠coq=∠eaq=60，

If OF=OC is intercepted on OE, △OCF is an equilateral triangle, ∴ Co = CF, ∠ OCF = ∠ ACE = 60,

So ∠ OCA+∠ QCF = ∠ FCE+∠ QCF, ∴∠ OCA = ∠ FCE, AC = EC, ∴△ OCA? △FCE(SAS)、

∴∠aoc=∠efc= 180-60= 120 ,∴∠aoe= 120-60=60，

Similarly, it can be proved that ∠ AOD = 60, ∴∠ AOD = ∠ AOE.

Using a four-point * * * circle is simple: ∴△DAC? △BAE(SAS),∴∠ACD=∠AEB,∴AECO four-point * * cycle,

∴∠ AOE =∠ ACE = 60, likewise ∠ AOD =∠ Abd = 60, ∴∠ AOD = ∠ AOE.

(3)∠ABC≦ADC diagram (3)

Even BD,∵BC=DC,∴∠CBD=∠CDB, if ab & gtAD, then ∠ ADB > ∠ABD，

∴∠adb+∠cdb>； ∠ABD+∠CBD, then ∠ ADC >; ∠ABC，

If ab

[diagonal AC bisects ∠BAD, which is of little significance for judging the quantitative relationship between ∠B and ∠D]