∠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