∠D=∠C=90°,∠DAB=∠ABC,若P为AB上一点,PM⊥BD于点M,PN⊥AC于点N,请猜想线段PM、PN、AD

PM+PN＝AD
证明：过点A作AG⊥MP,交MP延长线于点G
∵∠D=∠C=90,∠DAB=∠ABC,AB＝AB
∴△ABC全等于△BAD
∴∠CAB＝∠DBA
∵AG⊥MP,PM⊥BD
∴AG∥BD
∴∠GAP＝∠DBA
∴∠GAP＝∠CAB
∵AG⊥MP,PN⊥AC
∴PG＝PN
∴GM＝PG+PM＝PN+PM
∵∠D＝90,PM⊥BD,AG⊥MP
∴矩形AGMD
∴GM＝AD
∴PN+PM＝AD