摘要：22． 解:(1)同意．如图.设AD与EF交于点M. 由折叠知.∠BAD=∠CAD. ∠AME=∠AMF=90O． ------------------------------1分 ∴ 根据三角形内角和定理得 ∠AEF=∠AFE． ------------------------------------2分 ∴ △AEF是等腰三角形．···················································································· 3分 (2)图⑤中的大小是22．5o．··················································································· 4分

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