Question

一道竞赛题,在三角形ABC和三角形A1B1C1中,CD,C1D1分别是∠ACB和∠A1B1C1的角平分线,且CD=C1D

Answer 1

作AE‖BC,A1E1‖B1C1,连接DE,D1E1.∵∠AEC=∠DCB=∠DCA,∠A1E1C1=∠D1C1B1=∠D1C1A1,∴⊿AEC,⊿A1E1C1为等腰三角∵⊿AED∽⊿DBC,⊿A1E1D1∽⊿D1B1C1∴DE÷DC=AD÷DB,D1E1÷D1C1=A1D1÷D1B1∵CD=C1D1∴DE=D1E1∵AB=A1B1∴AD=A1D1,DB=D1B1∵∠ADC=∠A1D1C1∴⊿ADC≌⊿A1D1C1∴∠A=∠A1,∠ACD=∠A1C1D1,∵CD,C1D1为角平分线∴∠ACB=∠A1C1B1∵AB=A1B1∴⊿ABC≌⊿A1B1C1