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,AQ⊥BC,交EG与M,垂足为Q.求证：EM=

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