如图,已知等腰Rt△ABE和等腰Rt△ACD；∠BAE=∠CAD=90度,AM⊥DE于M,交BC于N,求证：AN为△AB

证明：过B作直线BP‖AC,并与MN交于P, ∠CBP=∠BCA;已知∠BAE=90°,∠AME=90°,∴∠BAP+∠EAM=∠EAM+AEM=90°,即∠BAP=∠AED△ABE与△ACD为等腰直角三角形,∴∠DAE=∠ACB+∠ABC=∠CBP+∠ABC=∠ABP,AB=AE∴△ABP≌△EAD,则AD=BP,又△ACD为等腰直角三角形,∴AC=AD=BP,BP‖AC,则∠NBP=∠NCA,∠NPN=NAC∴△NBP≌△NCA,得CN=NB∴AN为△ABC的中线证毕.