Diagonals of quadrilateral $ABCD$ intersect at $E$. Given $\frac{AE}{AC} = λ,\>\frac{BE}{BD} = μ$. Find ratios for sides

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

Solution 1:

$\color{blue}{(i)}$. 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} $$\implies [ABSR] = [ARP]+[ABSP] = \frac{\lambda\mu-\frac14}{\lambda+\mu-1} $$

Solution 2:

First remember these results:

1. Area ratio of similar triangles.

If $∆ABC \sim ∆PQR$ then $\frac{[∆ABC]}{[∆PQR]} = \frac{AB^2}{PQ^2}$.

2. Area ratio of triangles whose base is on the same line and have the same vertex. $$\frac{[∆PA_1B_1]} {[∆PA_2B_2]} = \frac{A_1B_1}{A_2B_2}$$
3. Area ratio of triangles having one common base and vertices are on the opposite side of the common base. $$\frac{[∆APB]}{[∆AQB]} = \frac{PX}{XQ}$$
4. Area ratio of triangles having one common base and vertices are on the same side of the common base. $$\frac{[∆APB]}{[∆AQB]} = \frac{PX}{XQ}$$ Corollary: If P, Q are two points on the same side of the line AB and R is a point on segment PQ such that $PR = k.PQ$ then $$[∆ABR] = (1-k)[∆ABP] + k[∆ABQ]$$ When $PQ || AB$ then the areas are equal, i.e. $[∆ABP] = [∆ABQ] = [∆ABR]$.
5. Area relation of triangles whose base and vertices are lying on a line. $$∆_2 = \frac{b}{m+n} (\frac{n}{a} ∆_1 + \frac{m}{c}∆_3)$$ When $a=b=c$ and $m = n$ then $∆_2 = \frac{1}{2}(∆_1+∆_3)$

The above results can be derived from :

1. Basic Area Lemma: If D is a point on the side BC of $∆ABC$ then $[∆ABD] : [∆ADC] = BD:DC$.
2. Intersection Lemma: If line segment PQ and RS intersect at point X then $PX:QX = [∆PRS]:[∆QRS]$.
3. Vertex Sliding Lemma: If the vertex of a triangle is moved then the resultant area is equal to the initial area times the ratio of final length to initial length of the vertex to base.

Now let me tell the algorithm that I followed to solve this question:

1. Convert the length ratio into area ratio using the area lemma. In the question above, $\frac{AR}{RD}$ can be written as $\frac{[∆ARP]}{[∆DRP]}$ or $\frac{[∆ARQ]}{[∆DRQ]}$.
2. Convert the area ratio into different area ratio using the results discussed above. In the given question, $[∆DRQ]$ can be written as $\frac{DQ}{EQ} [∆ERQ]$ using 1st result.
3. Convert the transformed area ratio into length ratio using results. In the question $\frac{[∆ARQ]}{[∆ERQ]}$ can be written as $\frac{AP}{EP}$ using results.

The key to find length ratio is to ask questions like: 'Which line defines point R on AD?' . After identifying the two line segments defining a point, apply intersection theorem to convert length ratio to area ratio. Then apply vertex Sliding Lemma to transform the area ratio(the trick is to identify the common base of the pair of triangles and then slide the vertex of any one of them (according to given data) to get a new ratio). Finally convert area ratio to length ratio.