Notation : We will use the Notation #A(bara)# to mention that the
position vector (pv) of a point #A" is "bara#.
Let, in the #DeltaABC, A(bara), B(barb), and, C(barc)#.
Given that, #P(barp)# is such that, #vec(PC)=3/2vec(BC)#.
Recall that, the pv of #vec(PC)=C(barc)-P(p)=barc-barp#.
#:. vec(PC)=3/2vec(BC)rArr barc-barp=3/2(barc-barb)#.
# :. 2barc-2barp=3barc-3barb#.
# rArr barp=1/2(3barb-barc)..................(1)#.
Similarly, if #Q(barq), and, C'(barc')#, then,
#vec(AQ)=1/4vec(AC) rArr barq=1/4(barc+3bara)......(2), and, #
clearly, #barc'=1/2(bara+barb)....................(3)#.
Now, prepare appropriate diagram for #DeltaABC#
#(AtoBtoC" anticlockwise", P" btwn. "B and C,#
# C'" btwn. "A and B, Q" btwn. "A and C")#.
In the said diagram, #PC'Q# is transverse in #DeltaABC.#
By Menelaus Theorem, if we can show that,
#(AC')/(C'B)(BP)/(PC)(CQ)/(QA)=1...(ast)," then "P,Q,C'" are collinear"#.
Here, #AC'=||vec(AC')||=||C'(barc')-A(bara)||=||1/2(bara+barb)-bara||#,
# rArr AC'=1/2||barb-bara||." Similarly, "C'B=1/2||barb-bara||#.
#BP=1/2||barb-barc||, PC=3/2||barc-barb||#.
#CQ=3/4||bara-barc||, QA=1/4||bara-barc||#.
Utilising these in #(ast)#, we have,
#(AC')/(C'B)(BP)/(PC)(CQ)/(QA)#,
#={(1/2||barb-bara||)/(1/2||barb-bara||)}{(1/2||barb-barc||)/(3/2||barc-barb||)}{(3/4||bara-barc||)/(1/4||bara-barc||)}#,
#={1}{1/3}{3/1}#
#rArr (AC')/(C'B)(BP)/(PC)(CQ)/(QA)=1#.
#"Therefore, "P,Q,C'" are collinear"#.
Enjoy Maths.!
