如图,已知AD是△ABC的外角∠EAC的平分线,AD与三角形ABC的外接圆相交于D,AC,BD相交与点P.

显示全部楼层 · 2012-12-1 21:11:46

证明：∵AD平分∠EAC∴∠CAD＝∠EAD∵∠CAD、∠CBD所对应圆弧都为劣弧CD∴∠CBD＝∠CAD∵四边形ABCD内接于圆∴∠BCD＝∠EAC∴∠CBD＝∠BCD∴BD＝CD∵∠ABD、∠ACD所对应圆弧都为劣弧AD∴∠ABD＝∠ACD∵∠APB＝∠DPC∴△ABP∽△DCP∴AB/PB＝CD/PC∴AB/PB＝BD/PC∴AB：BD＝PB：PC...

千问 · 2012-12-1 21:11:46

证明:连接CD.∵∠EAD=∠BCD;
∠DAC=∠CBD;
∠EAD=∠DAC.∴∠BCD=∠CBD,则BD=CD.又∵∠BAP=∠CDP;∠BPA=∠CPD.∴⊿BAP∽⊿CDP,AB:CD=PB:PC.故:AB:BD=PB:PC.(等量代换)...