Geometry At It's Best

We use 4 different ways to set triangle vertices with masses. On the picture each center of mass marked as O.

Well, $ar(PQRS) = ar(AFG) - ar(APR) - ar(QSGF)$

$ar(PQRS) = ar(AFG) - ar(APR) - ar(SGBD) + ar(DQFB)$ ----4

We will adopt the same strategy for finding the answer.

We know that

area of any triangle = $\frac{1}{2}absin\theta$

hence,

$ar(AFG) = \frac{1}{2}*ar(ABC)$

In ABF,

using Menealus' theorem, $\frac{BC}{CF}\frac{QF}{AQ}\frac{AD}{DB} = -1$

Hence, $\frac{QF}{AQ} = \frac{3}{8}, \frac{AF}{AQ} = \frac{11}{8}$

$\frac{ar(AFB)}{ar(ADQ)} = \frac{AB.AF}{AD.AQ} = \frac{33}{16}$

$\frac{ar(AFB)}{ar(DQFB)} = \frac{33}{17}$

Also, $ar(ABF) = \frac{1}{4}*ar(ABC)$

$\frac{ar(ABC)}{ar(DQFB)} = \frac{132}{17}$

$\frac{ar(DQFB)}{ar(ABC)} = \frac{17}{132}$ ----1

In BDC,

using Menealus' theorem, $\frac{AB}{AD}\frac{DS}{SC}\frac{GC}{BG} = -1$

hence, $\frac{DS}{SC} = 2, \frac{DC}{SC} = 3$

$\frac{ar(SGC)}{ar(DBC)} = 12$

$\frac{ar(DSGB)}{ar(DBC)} = \frac{11}{12}$

Also, $ar(DBC) = \frac{1}{3}*ar(ABC)$

$\frac{ar(DSGB)}{ar(ABC)} = \frac{11}{36}$ -----2

Similarly, $ar(APR) = \frac{4}{15}ar(AFG)$

$ar(AFG) - ar(APR) = \frac{11}{30}ar(ABC)$ -----3

Substituting 1, 2, 3 in 4,

$ar(PQRS) = \frac{94}{495}ar(ABC)$

$\frac{ar(ABC)}{ar(PQRS)} = \frac{495}{94}$