Question

如图,在梯形ABCD中,AD‖BC,∠B=90°,∠C=45°,AD=1,BC=5,E为AB中点,EF‖DC交BC于F,

FE的延长线交DA的延长于G，求BF，EF的长

Answer 1

过F作FG平行AD交AB于G，则G为AB中点。分别过E,G作EH,GK垂直BC于H,K，过A作AM垂直和胡EG于M,则EF=BK 设梯形的高为h 连AE,BE 因为E为DC的中点所以AM=（1/2)*h 在△AEG中，S=（1/2）*AM*EG=(1/2）*EF*AG=(1/搏棚陪2)*(1/2)*h*EG 在梯形ABCD中， S=S梯形ADEG S梯形BCEG =S△ADE S△AEG S△BGK S矩形EGKH S△CEH
梯形基蠢ABCD的面积=AB×EF =(1/2)*(1/2)*h*AD (1/2)*(1/2)*h*EG (1/2)*(1/2)*h*BK (1/2)*h*EG (1/2)*(1/2)*h*CH =(1/2)*(1/2)*h*EG (1/2)*(1/2)*h*(AD BK EG CH) (1/2)*(1/2)*h*EG =(1/2)*h*EG (1/2)*(1/2)*h*(AD BC) =(1/2)*h*EG (1/2)*h*EG =h*EG =2EF*AG =AB*EF