Question

如图,以三角形ABC的边ABAC为边向外作正方形ABDE和正方形ACFG，AH⊥BC，交EG与M，垂足为H。求证：EM=MG

Answer 1

分别过点E,G作EP⊥HM,GQ⊥HM垂足为P,Q
∵ABDE是正方形，∴AB=AE,∠BAE=90º,∴∠BAH+∠EAP=90º,∵AH⊥BC,∴∠BAH+∠ABH=90º∴∠ABH=∠EAP,∵∠AHB=∠P=90º∴⊿ABH≌⊿EPA,∴AH=EP,
同理可证AH=GQ,∴EP=GQ,易证⊿EPM≌⊿GQM,∴EM=GM,

分别过点E,G作EP⊥HM,GQ⊥HM垂足为P,Q

Answer 2

分别过点E,G作EP⊥HM,GQ⊥HM垂足为P,Q
∵ABDE是正方形，∴AB=AE,∠BAE=90º,∴∠BAH+∠EAP=90º,∵AH⊥BC,∴∠BAH+∠ABH=90º∴∠ABH=∠EAP,∵∠AHB=∠P=90º∴⊿ABH≌⊿EPA,∴AH=EP,
同理可证AH=GQ,∴EP=GQ,易证⊿EPM≌⊿GQM,∴EM=GM

Answer 3

分别过点E,G作EP⊥HM,GQ⊥HM垂足为P,Q
∵ABDE是正方形，∴AB=AE,∠BAE=90º,∴∠BAH+∠EAP=90º,∵AH⊥BC,∴∠BAH+∠ABH=90º∴∠ABH=∠EAP,∵∠AHB=∠P=90º∴⊿ABH≌⊿EPA,∴AH=EP,
同理可证AH=GQ,∴EP=GQ,易证⊿EPM≌⊿GQM,∴EM=GM,