Question

如图,四边形ABCD中,∠ADC=∠ABC=90°,AD=CD

Answer 1

作DF⊥BC于F；又DP⊥AB∴∠F＝∠DPB＝∠ABC=90°,∴四边形BPDF为矩形∴∠PDF=90°＝∠ADC∵∠ADP＝∠ADC－∠PDC＝∠PDF－∠PDC＝∠CDF,∠APD＝∠F＝90°,AD=CD∴⊿ADP≌⊿CDF(AAS)∴DP=DF∴矩形BPDF为正方形∴18＝四边形ABCD的面积＝S⊿ADP＋S四边形BPDC＝S⊿CDF＋S四边形BPDC＝S正方形BPDF＝DP²∴DP＝3√2