如图,∵AD=AB,∠DAB=90°,∴将△ADQ绕点A转90°得△ABQ',∴PQ'=PB+BQ'=PB+DQ,∵∠1=∠3,∠1=∠2,∴∠2=∠3,∴∠BAQ=∠Q'AP,∵AB∥CD,∴∠AQD=∠BAQ,又∵∠Q'=∠AQD,∴∠Q'=∠PAQ',∴PA=PQ'=PB+DQ (2)结论仍然成...