Question

P为△ABC内任一点 AP,BP,CP交BC,AC,AB于点Q,R,S 证 PQ/AQ+PR/BR+PS/CS=1

Answer 1

证明:作PM垂直BC于M,AN垂直BC于N.

则:PM∥AN,得:PM/AN=PQ/AQ;

S⊿PBC/S⊿ABC=(BC*PM/2)/(BC*AN/2)=PM/AN=PQ/AQ;(1)

同理;S⊿APC/S⊿ABC=PR/BR;(2)

S⊿APB/S⊿ABC=PS/CS.(3)

(1)+(2)+(3),得:(S⊿PBC+S⊿APC+S⊿APB)/S⊿ABC=PQ/AQ+PR/BR+PS/CS.

即:S⊿ABC/S⊿ABC=PQ/AQ+PR/BR+PS/CS.

故:PQ/AQ+PR/BR+PS/CS=1.