Let P, Q and R be three collinear points on BC, CA and AB of triangle ABC. Show that $BP.CQ. AR+PC. QA.RB=0$using section formula. Take proper sign for external section into account.
My try: I took P between B and C, Q between C and A, R outside of AB segment, so that P, Q, R are collinear. Then I took coordinates of A, B, C as $(x_1,y_1),(x_2,y_2), (x_3,y_3)$. Let BP:PC equals $l:1$, CQ:QA equals $m:1$,AB:BR equals $n:1$. Now I am stuck. Please help.
Let A, B, C be $(x_{1}, y_{1}), (x_{2}, y_{2}) ,(x_3, y_3)$ respectively
Let line L be ax+by=1
Let L divide sides in ratios l:1, m:1, n:1
Such that (BP)/(PC) = l/1, (CQ)/(QA) = m/1 and (AR)/(RB) = n/1
Therefore
$P=[ (lx_3 + x_2)/(l+1) , (ly_3 + y_2)/(l+1)]$,
$Q=[ (mx_1 + x_3)/(m+1) , (my_1 + y_3)/(m+1)], $
$R= [ (nx_2 + x_1)/(n+1) , (ny_2 + y_1)/(n+1)]$
since P, Q, R are in line L therefore they satisfy the equation of L Therefore putting them in equation of L we get
$1. l(ax_3 +by_3 -1)=-(ax_2 + by_2 - 1)$
$2. m(ax_1 +by_1 -1)=-(ax_3 + by_3 - 1)$
$3. n(ax_2 +by_2 -1)=-(ax_1 + by_1 - 1)$
Multiplying equations 1, 2 and 3 we get l * m * n = -1
That is {(BP)/(PC)} * {(CQ)/(QA)} * {(AR)/(RB)} = -1
With simple manipulation the statement in the question can be obtained