已知:如图,在Rt△ABC中,∠ACB=90°,BC的垂直平分线DE交BC于D,交AB于E,F是DE上一点.
(1)求证:四边形ACEF是平行四边形;
(2)当∠B的大小满足什么条件时,四边形ACEF是菱形?请回答并证明你的结论.
忘了打一个条件AF=CE
人气:194 ℃ 时间:2019-08-16 17:47:24
解答
(1)∵DE为BC的垂直平分线,
∴∠BED=∠CED,且DE‖AC,故E为AB的中点,
∴在Rt△ABC中,CE=AE=BE,
∴∠AEF=∠AFE,且∠BED=∠AEF,
∠DEC=∠DFA,
∴AF‖CE,
又∵AF=CE,
∴四边形ACEF为平行四边形;
(2)要使得平行四边形ACEF为菱形,则AC=CE即可,
∵CE=1/2 AB,
∴AC=1/2 AB即可,
在Rt△ABC中,∠ACB=90°,
∴当∠B=30°时,AB=2AC,
故∠B=30°时,四边形ACEF为菱形.
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