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

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Answer 1 · 2018-04-17T11:32:58

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.!