已知∠BCD=α∠BAD=β,CB=CD.(1)如图一,若α=β=90°,求证:AB+AD=根号已知∠BCD=α∠BAD=β,CB=CD. (1)如图一,若α=β=90°,求证:AB+AD=根号2AC (2)如图2, 若α=β=90°,求证:AB-AD= 根号2AC (3)如图3若α=120°
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![已知∠BCD=α∠BAD=β,CB=CD.(1)如图一,若α=β=90°,求证:AB+AD=根号已知∠BCD=α∠BAD=β,CB=CD. (1)如图一,若α=β=90°,求证:AB+AD=根号2AC (2)如图2, 若α=β=90°,求证:AB-AD= 根号2AC (3)如图3若α=120°](/uploads/image/z/6921222-6-2.jpg?t=%E5%B7%B2%E7%9F%A5%E2%88%A0BCD%3D%CE%B1%E2%88%A0BAD%3D%CE%B2%2CCB%3DCD.%EF%BC%881%EF%BC%89%E5%A6%82%E5%9B%BE%E4%B8%80%2C%E8%8B%A5%CE%B1%3D%CE%B2%3D90%C2%B0%2C%E6%B1%82%E8%AF%81%EF%BC%9AAB%2BAD%3D%E6%A0%B9%E5%8F%B7%E5%B7%B2%E7%9F%A5%E2%88%A0BCD%3D%CE%B1%E2%88%A0BAD%3D%CE%B2%2CCB%3DCD.+%EF%BC%881%EF%BC%89%E5%A6%82%E5%9B%BE%E4%B8%80%2C%E8%8B%A5%CE%B1%3D%CE%B2%3D90%C2%B0%2C%E6%B1%82%E8%AF%81%EF%BC%9AAB%2BAD%3D%E6%A0%B9%E5%8F%B72AC+%EF%BC%882%EF%BC%89%E5%A6%82%E5%9B%BE2%2C+%E8%8B%A5%CE%B1%3D%CE%B2%3D90%C2%B0%2C%E6%B1%82%E8%AF%81%3AAB-AD%3D+%E6%A0%B9%E5%8F%B72AC+%EF%BC%883%EF%BC%89%E5%A6%82%E5%9B%BE3%E8%8B%A5%CE%B1%3D120%C2%B0)
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