Help with this problem involving a triangle

Question

The area of the triangle is 85 square centimetres, the side $AB$ is divided by the point $P$ so that the ratio $BP:AP$ is 1:4 and the side $AC$ is divided by $Q$ so that the ratio $AQ:CQ$ is 2:3. Find the area of $\triangle BXC$ and the ratios $BX:XQ$ and $CX:XP$.

I have found that $\triangle AQB$ = 34, $\triangle BXC$ = 68, $\triangle QCB$ = 51 and $\triangle PBC$ = 17. However, I cannot find a way to determine the area of $\triangle BXC$ with the information that I have found

Answer 1

Let $CX:XP=s:(1-s)$.

$[\triangle BCP]=85\div 5=17$. Hence $[\triangle BXC]=17s$ and $[\triangle PXB]=17(1-s)$.

$\displaystyle [\triangle CPQ]=\frac{3}{5}[\triangle CAP]=\frac{3}{5}\times\frac{4}{5}\times85=\frac{204}{5}$.

$\displaystyle [\triangle QXP]=(1-s)[\triangle CPQ]=\frac{204(1-s)}{5}$.

$\displaystyle [\triangle BPQ]=\frac{1}{5}[\triangle BAQ]=\frac{1}{5}\times\frac{2}{5}\times 85=\frac{34}{5}$.

\begin{align*} 17(1-s)+\frac{204(1-s)}{5}&=\frac{34}{5}\\ s&=\frac{15}{17} \end{align*}

$CX:XP=15:2$.

$\displaystyle [\triangle BXC]=17s=15$.

$\displaystyle [\triangle QXC]=85\times \frac{3}{5}-15=36$.

$BX:XQ=85:370=15:36=5:12$.

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Alternatively, by Menelaus Theorem,

$\displaystyle \frac{BP}{PA}\times\frac{AC}{CQ}\times\frac{QX}{XB}=1$ and $\displaystyle \frac{CQ}{QA}\times\frac{AB}{BP}\times \frac{PX}{XC}=1$.

Answer 2

Do the same problem for the triangle with vertices $A=(0,0)$, $B=(5,0)$, $C=(0,5)$ first. All relevant ratios will be the same.

By the way: It is unclear from the text whether $P$ is nearer to $A$ or to $B$; similarly for $Q$.

Answer 3

You can use ladder theorem all these represented quantities are areas.

$$\frac{1}{\triangle ABC}+\frac{1}{\triangle XBC}=\frac{1}{\triangle PXB+\triangle XBC}+\frac{1}{\triangle XQC+ \triangle XBC}$$ $\triangle ABC=85$ $\triangle ABQ=34$ $\triangle CBQ=51$ $\triangle APC = 17$ $\triangle PBC=68$

let the area of quadrilateral APXQ be $d$ $$ar. \triangle PBX = \triangle ABQ-d$$ $$ar. \triangle XQC =\triangle APC -d$$ $$ar. \triangle PBX = 34-d$$ $$ar. \triangle XQC = 17-d$$ Now plug in the values $$\frac{1}{85}+\frac{1}{\triangle XBC}=\frac{1}{34-d+\triangle XBC}+\frac{1}{17-d+ \triangle XBC}$$ one more equation can be formed i.e. $$\triangle ABC =d+34-d+17-d+\triangle XBC$$ $$ 14 +d=\triangle XBC$$ put this in above equation $$\frac{1}{14 +d}=\frac{1}{48}+\frac{1}{31 }-\frac{1}{85}$$ by solving $d=10.197$ and $\triangle XBC =24.197$