题目内容
如图,在矩形ABCD中,AB=2BC,在CD上取一点E,使AE=AB,则∠EBC的度数为( )
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A.30° | B.15° | C.45° | D.不能确定 |
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作EF⊥AB于F,则EF=BC,
又∵AB=2BC,AE=AB,
∴AE=2EF,
∴∠EAF=30°,
∵AE=AB
∴∠ABE=∠AEB=75°,
∴∠EBC=90°-75°=15°.
故选B.
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又∵AB=2BC,AE=AB,
∴AE=2EF,
∴∠EAF=30°,
∵AE=AB
∴∠ABE=∠AEB=75°,
∴∠EBC=90°-75°=15°.
故选B.
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