What's the measure of the segment $QC$ in the triangle below?

Question

For reference: In a triangle $ABC$ with centroid $G$, on $AB$ and $BC$ the points $P$ and $Q$ are located respectively such that $BP = 4AP$ and $BQ = 8$. Calculate $QC$, If $P, G ~and~ Q$ are collinear.(answer: $\frac{16}{3}$)

Solution by geometry.

My progress:

I do not have many ideas in this exercise

Try Menelaus: $\triangle CPQ-BD\\ CI.PG.8 = IP.GQ.(8+QC)\\ \triangle ABD-PC\\ x.BI.CD=4x.DI.AC\implies BI = 8DI\\ BI = BG+GI ~and ~DG = GI+DI\implies DI = DG-GI\\ BG+GI = 8(GI+DI)\rightarrow 2GD +GI = 8GD -8GI \implies GI = \frac{2GD}{3}\\ $

but I did not see much utility...

Answer 1

Let $D'$ be the intersection of lines $PGQ\cap ADC$. Menelaus for $\Delta BAD$ w.r.t. the transversal line $PGD'$ gives $$ 1= \frac{PA}{PB}\cdot \frac{GB}{GD}\cdot \frac{D'D}{D'A} = \left(-\frac 14\right)\cdot \left(-2\right)\cdot \frac{D'D}{D'A} =\frac 12\ . $$ This shows which is the location of the point $D'$ on $AC$, it is so that $$ D'A=AD=DC\ . $$ A faithful picture is needed. (Please compare with the given picture...)

Now apply Menelaus in $\Delta BAC$ w.r.t. the transversal $D'PQ$ to get $$ 1 = \frac{D'A}{D'C}\cdot \frac{QC}{QB}\cdot \frac{PB}{PA} = \frac 13\cdot \frac{QC}{QB}\cdot (-4) \ , $$ which clears the location of the point $Q$ on $BC$.

$\square$

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Here is one more picture, showing that when $BQ$ is realized as "four pieces", then $QC$ is "three (same) pieces."

Answer 2

First apply Menelaus's Theorem in $\triangle ACP$ with traversal $DB$.

$\displaystyle \frac{AB}{BP} \cdot \frac{PI}{IC} \cdot \frac{CD}{DA} = 1$

Find relation between $IP$ and $IC$

Next, apply Menelaus's in $\triangle ABD$ with traversal $PC$.

$\displaystyle \frac{AC}{CD} \cdot \frac{DI}{IB} \cdot \frac{BP}{PA} = 1$

Using $DI = DG - IG, IB =IG + GB$, find relation between $IG$ and $GB$

Finally, apply Menelaus's in $\triangle BIC$ with traversal $QP$

$\displaystyle \frac{CP}{IP} \cdot \frac{IG}{GB} \cdot \frac{BQ}{QC} = 1$

and that gives you value of $QC$.