几何线段平分证明有任意三角形ABE,分别以AE,AB为直角边,构造等腰直角三角形RtΔADE,RtΔABC,以BE为斜边

连接GH,GJ,AH,AJ,则∠GEH=∠GEB+∠BEH=45°+∠BEH
=∠AED+∠BEH=∠AEB,而HE/AE=1/√2=GE/BE
∴△ABE∽△HGE,同理有∠GBJ=∠EBA,GB/BE=BJ/AB
∴△ABE∽△JBG,∴△HGE∽△JBG,又GE=BG
∴△HGE≌△JBG,∴GJ=EH=AH,GH=BJ=AJ
又HJ=HJ,∴△AHJ≌△GJH,∴∠AJH=∠GHJ,∠AHJ=∠GJH
即AJ//HG,AH//GJ,即AHGJ是平行四边形,∴AG和HJ互相平分