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證明:過B點作BF⊥BC交CE的延長線於F, 很明顯△CBF≌△ACD ∴∠F=∠CDA;CD=BF ∵CD=BD∴BD=BF ∵∠ABC=∠EBF=45度;BE=BE ∴△DBE≌△FBE ∴∠EDB=∠F ∴∠CDA=∠EDB.
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