Concurrency at interior point

pankaj · #1 $Old$ January 13th, 2009, 11:40 AM

Let $P$ be an interior point of $\triangle ABC$ such that the lines $AA_{1},BB_{1},CC_{1}$ are concurrent at $P$ and the points $A_{1},B_{1},C_{1}$ lie on $BC,CA,AB$ respectively.Find using vectors or otherwise value of
$\frac{PA_{1}}{AA_{1}}+\frac{PB_{1}}{BB_{1}}+\frac{PC_{1}}{CC_{1}}$

red_dog · #2 $Old$ January 14th, 2009, 03:10 AM

Let $PQ\perp BC, \ AD\perp BC$

Then, $\frac{PA_1}{AA_1}=\frac{PQ}{AD}$

$A_{\Delta PBC}=\frac{PQ\cdot BC}{2}, \ A_{\Delta ABC}=\frac{AD\cdot BC}{2}$

$\frac{A_{\Delta PBC}}{A_{\Delta ABC}}=\frac{\frac{PQ\cdot BC}{2}}{\frac{AD\cdot BC}{2}}=\frac{PQ}{AD}$

So, $\frac{PA_1}{AA_1}=\frac{A_{\Delta PBC}}{A_{\Delta ABC}}$

In a similar way we have

$\frac{PB_1}{BB_1}=\frac{A_{\Delta PAC}}{A_{\Delta ABC}}$ and $\frac{PC_1}{CC_1}=\frac{A_{\Delta PAB}}{A_{\Delta ABC}}$

Then $\frac{PA_1}{AA_1}+\frac{PB_1}{BB_1}+\frac{PC_1}{CC_1}=\frac{A_{\Delta PBC}+A_{\Delta PAC}+A_{\Delta PAB}}{A_{\Delta ABC}}=\frac{A_{\Delta ABC}}{A_{\Delta ABC}}=1$