Question

正方形ABCD的对角线交点为O,AE平分∠BAC交BC于E,交OB于F求证:EC=2FO

Answer 1

证明：过E点做DH⊥AC

∵AE平分∠BAC EB⊥AB

根据角平分线定理

∴BE=EF

设bc=a

AC=√2a

AO=√2/2 a

EC=√2EH=√2BE

EC+BE=BC

（1+√2）BE=a

BE=（√2-1）a=HC

AH=AC-HC=a

∵正方形的对角线互相垂直平分

∴OF‖EH

∴AO/AH=OF/EH

√2/2 a /a =OF/√2/2EC

OF/EC=√2/2*√2/2=1/2

EC=2FO