题目内容
如图,等腰梯形ABCD中,AD∥BC,AD=AB=5,且∠ABC=45°,则BC等于______.
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过A作AE⊥BC于E,过D作DF⊥BC于F,
则AE∥DF,∠AEB=∠DFC=∠AEF=90°,
∵AD∥BC,
∴四边形AEFD是矩形,
∴AD=EF=5,AE=DF,
在Rt△AEB和Rt△DFC中
|
∴Rt△AEB≌Rt△DFC(HL),
∴BE=CF,
在Rt△AEB中,cos45°=
BE |
AB |
BE |
5 |
∴BE=
5 |
2 |
2 |
即CF=
5 |
2 |
2 |
∴BC=BE+EF+CF=2×
5 |
2 |
2 |
2 |
故答案为:5+5
2 |
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