如图,等腰三角形ABC,等腰三角形ADE,∠BAC=∠DAE=90°,p是CD中点,连接BE,试探索AP与BE的关系
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如图,等腰三角形ABC,等腰三角形ADE,∠BAC=∠DAE=90°,p是CD中点,连接BE,试探索AP与BE的关系(图2)延长AP,使PF=AP,连接DF,做DG⊥AC交于点G,AP与BE交点为O
∠EAG=∠DAE-∠DAG=90-∠DAG ∠AD=∠BAC-∠DAG=90-∠DAG
∠EAG=∠BAD DG//AB ∠BAD=∠ADG ∠GAE=∠ADG
AP=PF ∠APC=∠DPF DP=CP △APC≌△FPC
AC=DF AB=AC AB=DF ∠PAC=∠PFC AC//DF∠GDF=90
∠ADF=∠ADG+∠GDF=∠ADG+90 ∠EAB=∠EAG+BAG=∠EAG+90
∠EAB=∠ADF AE=AD
△EAB≌△ADF BE=AF AF=2AP BE=2AP
∠ABE=∠AFD ∠PAC=∠PFC ∠ABE=∠PAC
∠ABE+∠BAO=∠PAC+∠BAO=∠BAC=90
∠AOB=180-(∠ABE+∠BAO)=90 BE⊥AP
结论:BE=2AP BE⊥AP