题目

如图,矩形ABCD中,BC=2AB=4,AE平分∠BAD交边BC于点E,∠AEC的分线交AD于点F,以点D为圆心,DF为半径画圆弧交边CD于点G,则的长为________ 答案:【答案】(2﹣)π.【解析】先由矩形的性质得出,∠BAD=∠B=∠D=90°,AD=BC=4,AD∥BC,根据AE平分∠BAD得到∠BAE=∠EAD=45°,那么△ABE是等腰直角三角形,于是AB=BE=2,AE=AB=2 . 再由∠AEC的分线交AD于点F,∠AEF=∠CEF,由AD∥BC,得出∠CEF=∠AFE,等量代换得到∠AEF=∠AFE,那么AF=AE=2,DF=AD﹣AF=4﹣2 ,然后根据弧It is not a serious illness. I guess she will_____ soon. A.release           B.recover            C.ruin              D.rescue
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