P是三角形ABC内一点,延长AP、BP、CP与对边相交,图中a、?
S△PBC/S△ABC=PD/AD=d/(a+d), 同理S△PCA/S△ABC=d/(b+d), S△PAB/S△ABC=d/(c+d), 三式相加得1=d[1/(a+d)+1/(b+d)+1/(c+d)], a+b+c=43,d=3, ∴(3+a)(3+b)(3+c)=3[(3+b)(3+c)+(3+c)(3+a)+(3+a)(3+b)], ∴27+9(a+b+c)+3(ab+bc+ca)+abc =3[27+6(a+b+c)+bc+ca+ab)], ∴abc=54+9*43=441.