In the plane of the triangle #ABC# we have #P# and #Q# such that #vec(PC)=3/2vec(BC)#and #vec(AQ)=1/4vec(AC)#,and #C'#is the middle of #[AB]#.How to demonstrate that #P,Q # and #C'# are collinear points with theorem of Menelaus?

Question

Sadie Allen · Accepted Answer

Please refer to a Proof in the Explanation.

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.!