已知凸五边形AXBCZ,点Y在ΔABC内,点P在BC上,∠YCB=∠XAB,∠PYC=∠AXB,∠YBC=∠CAZ,∠BYP=∠AZC,求证:X、Y、Z三点共线,且XY/YZ=BP/CP。 证明:用复数方法。由ΔYBP∽ΔZAC得(Y-P)/(B-P)=(Z-C)/(A-C);ΔYCP∽ΔXAB,得(Y-P)/(C-P)=(X-B)/(A-B)。由第一式得Y(A-C)-PA=Z(B-P)-BC,由第二式得Y(A-B)-PA=X(C-P)-BC,两式相减得Y(B-C)=Y[(B-P)-(C-P)]=Z(B-P)-X(C-P),即(Y-Z)(B-P)=(Y-X)(C-P),(B-P)/(P-C)=(X-Y)/Y-Z),因B、P、C共点,故X、Y、Z共点,且XY/YZ=BP/PC。 |
|