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證明:作BG⊥AE於G,∵四邊形ABCD是正方形,DF⊥AE,∴∠AFD=∠AGB=90°,∵∠DAF+∠GAB=90°,∠DAF+∠ADF=90°,∴∠ADF=∠GAB,又AD=AB,∴△ADF≌△BAG.
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