Question

已知p为三角形ABC内一点,且3AP+4BP+5CP=0,延长AP,交BC于D,若AB=a,AC=b,试用a,b表示AD

Answer 1

AP-BP=AB

AP-CP=AC

3AP+4(AP-AB)+5(AP-AC)=12AP-4AB-5AC=0

AP=AB/3+5AC/12

AD=hAP=hAB/3+5hAC/12

BD=-AB+AD=-AB+hAB/3+5hAC/12=(h-3)AB/3+5hAB/12

BD=xBC=x(AC-AB)=xAC-xAB

x=(h-3)/3

x=-5h/12

h=4/3

AD=4AB/9+5AC/9