In the figure $ABCD$ is a convex quadrilateral. its diagonals $AC$ and $BD$ intersect at $E$, and their midpoints are $P$ and $ Q$, respectively. Given that $\frac{AE}{AC} = λ,\>\frac{BE}{BD} = μ$, (i) Find the ratios $\frac{AR}{RD}$ and $\frac{BS }{SC }$ in terms of λ and μ. (ii) Suppose the area of $ABCD$ is 1, what is the area of $ABSR$?
The attached below contains all my work. I know it is not very neat but I don't think it would matter anyway as I couldn't make any meaningful process either 🙁
Also, in my book, I am struggling a lot in the topic 'Area lemma' (this question is from that topic) , so is there any suggestion / algorithmic way to approach these problems.
Thanks!
PS; please don't use any fancy stuff (like complex / bary). A solution with Menelaus, Ceva's, Area lemma would be ideal.
EDIT ; I figured out the first part, menelaus on ADE with RS transversal
Best Answer
$\color{blue}{(i)}$. Given the midpoints $P$ and $Q$, establish
$$ \frac{EP}{AP} = \frac{AE-AP}{AP} =2\lambda-1,\>\>\>\>\>\frac{EQ}{DQ} = \frac{BE-BQ}{DQ} =2\mu-1 $$
With [.] denoting areas, derive the ratio $\frac{AR}{RD}$ below and, likewise, $\frac{BS }{SC}$
$$x=\frac{AR}{RD} = \frac{[ARQ]}{[DRQ]} = \frac{[ARQ]}{\frac{DQ}{EQ} [ERQ]} = \frac{EQ}{DQ} \frac{AP}{EP}=\frac{2\mu-1}{2\lambda-1} $$ $$y=\frac{BS }{SC}=\frac{2\lambda-1}{2\mu-1}$$
$\color{blue}{(ii)}$. Note $[ABC]=\mu[ABCD]=\mu$ and $[ADC]=(1-\mu)[ABCD]=1-\mu$ \begin{align} &[ARP]= \frac x{1+x}[ADP] =\frac x{1+x} \frac12[ADC] = \frac {\frac12x(1-\mu)}{1+x}\\ &[ABSP] = [ABC] - \frac12[ASC] = [ABC] - \frac{\frac12}{1+y}[ABC] =\frac {(y+\frac12)\mu}{1+y} \end{align} $$\implies [ABSR] = [ARP]+[ABSP] = \frac{\lambda\mu-\frac14}{\lambda+\mu-1} $$