In $\Delta ABC$, $D,E,F$ are points on the sides $BC,CA,AB$ respectively, such that the lines joining them are concurrent at a point $G$.

Question

In $\Delta ABC$, $D,E,F$ are points on the sides $BC,AB,CA$ respectively, such that the lines joining them are concurrent at a point $G$, and $BD = 2CD$. Let $[\Delta GEC] = 3$, $[\Delta GCD] = 4$. Find $[\Delta ABC]$.

What I Tried: Here is a picture :-

At this point I was only able to figure out that [$\Delta GBD] = 8$. I have been only given that $CD = x$, $BD = 2x$ . I have no other idea on how to use similarity or area relations here, although I think the only way to do this problem is by that method. I also think Ceva's Theorem might work, but how? We don't know any lengths of sides $BE,EA,AF,FC$ and so on.

Can anyone help me? Thank you.

Answer 1

The ratio between area of two triangles having same height but different base equals the ratio between their length of bases

$$\frac{FG}{BG}=\frac{[\triangle AFG]}{[\triangle AGB]}= \frac{[\triangle CFG]}{[\triangle CBG]} = \frac{3}{12}=\frac{1}{4}$$

$$\frac{CD}{BD}=\frac{[\triangle ACD]}{[\triangle ABD]}=\frac{7+[\triangle AFG]}{8+[\triangle AGB]}=\frac{1}{2}$$

$$2(7+[\triangle AFG])=8+[\triangle AGB]$$

$$6+2[\triangle AFG]=[\triangle AGB]$$

$$\frac{[\triangle AFG]}{6+2[\triangle AFG]}=\frac{1}{4}$$

$$[\triangle AFG]=3,[\triangle AGB]=12$$

Hence $[\triangle ABC]=30$

Answer 2

Let $S_{\Delta AGF}=a$, $S_{\Delta AGE}=b$ and $S_{\Delta BGE}=c$.

Thus, $$\frac{a+3+4}{b+c+8}=\frac{4}{8},$$ which gives $$b+c=2a+6.$$ Also, $$\frac{a+b+c}{8+4+3}=\frac{a}{3},$$ which gives $$b+c=4a.$$ Thus, $$4a=2a+6,$$ $$a=3,$$ which says that $BF$ is a median of $\Delta ABC.$

Id est, $$S_{\Delta ABC}=2S_{\Delta BCF}=2\cdot15=30.$$