在正方形ABCD中,E是BC延长线上的任意一点,连接AE,作AE的垂线与正方形的外角平分线相交于点F,AE=EF?

Question

在正方形ABCD中,E是BC延长线上的任意一点,连接AE,作AE的垂线与正方形的外角平分线相交于点F,AE=EF?

Answer 1

设AG=EP=BH=x;AE=GP=DF=y;AB=BC=CD=AD=a;
逆时针旋转△AFD,使AD与AB重合；F点落于点K;
∵△AFD全等于△AKB;
∴AK=AF；∠KAB=∠FAD;∠ABK=∠ADF=90°;BK=DF=y;
∠ABK+∠ABC=90+90=180°；K,B,C共线；
S△AKH=1/2(BK+BH)*AB=1/2a(x+y);
S△AFH=S正方形ABCD-S△ABH-S△ADF-S△CHF
S正方形ABCD=a^2;
S△ABH=1/2ax;
S△ADF=1/2ay;
S△CHF=1/2(a-x)(a-y)
∴S△AFH=a^2-1/2ax-1/2ay-1/2(a-x)(a-y)=1/2a^2-1/2xy;----(1)
∵S矩形PHCF=（a-x)(a-y);
S矩形AEPG=xy;
S矩形PHCF=2S矩形AEPG;
∴（a-x)(a-y)=2xy;
1/2xy=1/2a^2-1/2a(x+y);----(2);
将（2）代入（1)得：
S△AFH=1/2a(x+y);
∴S△AKH=S△AFH;
∵S△AKH=1/2AK*AHsin∠KAH;
S△AFH=1/2AF*AHsin∠HAF;
∴1/2AK*AHsin∠KAH=1/2AF*AHsin∠HAF;
AK=AF;
∴∠HAF=∠KAH=1/2∠KAF;
∵∠BAF+∠FAD=90°∠FAD=∠KAB;
∴∠KAB+∠BAF=∠KAF=90°；
∴∠HAF=1/2*90°=45°
结论：角HAF的度数为45°