An Olympiad geometry question on area lemma.

Let [.] denote areas. Evaluate the area ratio

$$\frac{[CBM]}{[EBM]}= \frac{[CBG]}{[EBG]} = \frac{\frac 14 [ABC]}{\frac 34 [ABG]}= \frac{\frac 14 [ABC]}{\frac 34 \frac 34[ABC]} =\frac 49 $$

which leads to the area $[CBM]$, and likewise $[ACK]$, $[BAL]$

$$[CBM] = \frac{4}{4+9} [CEB] = \frac{4}{13}\frac 34[ABC] =\frac{3}{13} = [ACK] = [BAL] $$ Then $$[KLM] = [ABC] -[CBM]- [ACK]-[BAL] =1-3\cdot\frac{3}{13}=\frac 4{13}$$

---

The derivation applies to general ratio $m:n$ as well $$\frac{[CBM]}{[EBM]}=\frac{m(m+n)}{n^2}\implies \frac{[CBM]}{[CBE]}=\frac{m(m+n)}{m(m+n)+n^2} $$ $$\implies [CBM]= \frac{m(m+n)}{m(m+n)+n^2}\cdot\frac{n}{m+n} =\frac{m n}{m^2+mn +n^2}$$ $$[KLM] = 1-3[CBM]= \frac{(m-n)^2}{m^2+mn +n^2} $$

The general case of this problem is called Routh's theorem. It states that if $EB/AE = x$, $FC/BF = y$, $GA/CG = z$, then $$\frac{[KLM]}{[ABC]} = \frac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)}.$$ Note I have taken the reciprocals so that in your case, $x = y = z = 3$. The proof of this theorem is given in the Wikipedia link.