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三角形ABC,BC邊上的中線AD,角B是角C的三倍.且角ADB為45°,求證ABC為直角三角形即證3C+C=90做DE垂直BC交AC於E,連接BE,過A做AF垂直DE,AG垂直BC易得:BE= EC EBC=C ABE=3C-C=2C=BEA AB=AEAGD等腰直角三角形AGD AFDG正…
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