A Pentagon AFBCGA is inscribed in a Triangle ∆ABC , what will be it's area?

Question

Consider a $∆ABC$ and a Pentagon $AFBCGA$ as shown below:
Here all the dotted red lines are perpendicular to a common side $BC$, their lengths are also given with red pen ; also length of $BC$ is given.Now how will we calculate the area of the inscribed Pentagon?

My thoughts:
If we could find $BM,MN,NP,PQ,QR \, \& \, RC$;
Then by using area of right triangle and area of trapezium formulas ,we can find the area of the required Pentagon

My Approach:
Let $BM=f,MN=e,NP=d,PQ=c,QR=b \, \& \, RC=a$;
then, $$a+b+c+d+e+f=10 $$ and by using similar triangle concepts we will get $5$ more independent equations on $a,b,c,d,e \,\& \,f$
i.e, In $∆ABP $ $$\frac{5}{8}=\frac{f}{f+e+d} \quad \quad \because ∆DBM \sim ∆ABP$$ and so on...

So, total six variables and six independent equations, hence we might expect an unique solution after solving this set of simultaneous linear equations.

But, Wolfram app is giving it's coefficient matrix determinant as $ 0$.
Hence,they are not all independent
So how to solve this problem?Any help please...

Answer 1

There is no need to deal with that many variables. Notice $$\frac{\verb/Ar/(ABF)}{\verb/Ar/(ABC)} = \frac{DF}{DC} = 1 - \frac{FC}{DC} = 1 - \frac{FN}{DM} = 1 - \frac45 = \frac15$$ $$\frac{\verb/Ar/(CAG)}{\verb/Ar/(CAB)} = \frac{EG}{EB} = 1 - \frac{GB}{EB} = 1 - \frac{GQ}{ER} = 1 - \frac23 = \frac13$$

We have $$\begin{align} \verb/Ar/(AFBCG) &= \verb/Ar/(ABC) - \verb/Ar/(ABF) - \verb/Ar/(CAG)\\ &= \left(1 - \frac15 - \frac13\right)\verb/Ar/(ABC)\\ &= \frac7{15} \times \frac{(AP)(BC)}{2}\\ &= \frac{56}{3}\end{align}$$

Answer 2

I considered BM=a, MN=b, NP=c, PQ=d, QR=e and RC=f. Then by the same method as you told I got the following six equations 4b-c-d-e-f=0 5f-3d-3e=0 3a-5b-5c=0 a+b+c+d+e+f=10 2b+2c+2d-3e-3f=0, and c+d-e-f=0. On solving these six equations we get unique values of all the variables i.e. a=55/12 , b=13/12 , c=5/3 , d=1/2 , e=7/6 , f=1.