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20 13 mathematical answers for the senior high school entrance examination in Beijing
analyse

(2) Connecting PC and AD is easy to prove.

∴ap=pc∠ ADB = ∠ CDB ∠PAD=∠PCD.

PQ = PA

∴pq=pc,∠adc=2∠cdb,∠pqc=∠pcd=∠pad

So ∠ pad+∠ PQD = ∠ PQC+∠ PQD = 180.

So ∠ apq+∠ ADC = 360-(∠ pad+∠ pqd) =180.

∴∠ADC= 180 -∠APQ= 180

∴2∠CDB= 180

∴∠CDB=90

(3)CDB = 90,PQ=QD

∴∠pad=∠pcq=∠pqc=2∠cdb= 180

Point p does not coincide with point b and point m.

∴∠bad>; ∠PAD & gt; Crazy

therefore