...点N在AC边上,并且角MDN=90度,若BM^2+CN^2=DM^2+DN^2.答:延长ND使DE=ND,连接BE,ME.BD=CD, ∠BDE=∠CDN, ND=ED △BDE≌△CDE BE=CN ∠DBE=∠DCN, BE//AC DE=ND, MD⊥NE MN=ME DM^2+DN^2=MN^2=ME^2=BM^2+CN^2 BE=CN ME^2=MB^2+BE^2 ∠MBE=90, BE//AC ∠BAC=90, △BAC为Rt△ AD=BC/2 BC^2=AB^2+AC^2 AD^2...