Finding the area of inner triangle constructed by three cevian lines of a large triangle

Question

QUESTION:

In a triangle $ABC$, $AD, BE$ and $CF$ are three cevian lines such that $BD:DC = CE:AC = AF:FB = 3:1$. The area of $\triangle ABC$ is $100$ unit$^2$. Find the area of $\triangle HIG$

Attempt:

First, I thought that it can be done through applying Menelaus's Theorem. Despite having few knowledge about that theorem, I started like that:

As a consequence of Menelaus's theorem, if two cevian $BH$ and $GF$ of $ABG$ meet at $I$ then:

$$\frac{HI}{IB} = \frac{BF}{AB}.\frac{AH}{HG}$$

$\implies \frac{HI}{IB} = \frac{AH}{4HG}......(i)$

And

$$\frac{FI}{IG} = \frac{3HG}{AG}......(ii)$$

Similarly from $\triangle ACI$, I got two more equations and that is:

$\frac{CG}{GI} = \frac{4GH}{AE}.......(iii)$

$\frac{EH}{HI} = \frac{IG}{4IC}........(iv)$

And likewise, from $\triangle BHC$,

$\frac{DG}{HG} = \frac{3HI}{BH}........(v)$

$\frac{IG}{GC} = \frac{BI}{4IH}.........(vi)$

But later on, I thought that it wouldn't be so good for relating with so many sides or segments and also be ineffective instead of relating the area. Then, how could I relate the area using Menelaus's theorem or in any other way like pure geometry?

Thanks in advance.

Answer 1

Let $N\in EC$ such that $DN||BE$ and $NC=y$.

Thus, by Thales $EN=3y$ and $AE=\frac{1}{3}EC=\frac{4}{3}y,$

which says $$\frac{AH}{HD}=\frac{AE}{EN}=\frac{\frac{4}{3}y}{3y}=\frac{4}{9}.$$ Also, let $M\in DC$ such that $EM||AD$ and $DM=x$.

Thus, by Thales again we obtain: $MC=3x$, $BD=12x$ and $$\frac{BH}{HE}=\frac{BD}{DM}=\frac{12x}{x}=12.$$

Similarly, $$AG:GD=CI:IF=12:1$$ and $$BI:IE=CG:GF=4:9,$$ which gives $$AH:HG:GD=BI:IH:HE=CG:GI:IF=4:8:1.$$ Now, let $S_{\Delta HIG}=s$.

Thus, $$\frac{S_{\Delta GFA}}{s}=\frac{9\cdot12}{8\cdot8}=\frac{27}{16},$$ which gives $$S_{\Delta GFA}=\frac{27}{16}s.$$ Also, $$\frac{S_{\Delta AFC}}{\frac{27}{16}s}=\frac{FC}{FG}=\frac{13}{9},$$ which gives $$S_{\Delta AFC}=\frac{39}{16}s$$ and since $$\frac{S_{\Delta ABC}}{\frac{39}{16}s}=\frac{4}{3},$$ we obtain: $$S_{\Delta ABC}=\frac{13}{4}s$$ and $$s=\frac{400}{13}.$$

Answer 2

I found a solution but actually it's not mine. I found it on a book and noticed that it is quite different from the Routh's theorem and it has been solved by the pure geometic way. And that is:

Point $C$ and $H$ are connected.

$\frac{\triangle ADC}{\triangle ADB} = \frac{1}{3}$ and $\frac{\triangle HDC}{\triangle HDB} = \frac{1}{3}$

Hence, $$\frac{\triangle ACH}{\triangle ABH} = \frac{1}{3}$$.

Similarly, $$\frac{\triangle CHB}{\triangle ABH} = 3$$

Now, $\triangle ABC = \triangle ABH + \triangle ACH +\triangle BHC = \triangle ABH + \frac{1}{3}\triangle ABH + \triangle ABH = \frac{13}{3} \triangle ABH$

So, $$\triangle ABH = \frac{3}{13} \triangle ABC$$

Similarly, $$\triangle ACG = \frac{3}{13} \triangle ABC$$

And, $$\triangle BIC = \frac{3}{13} \triangle ABC$$

Therfore, $\triangle HIG = \triangle ABC - \triangle ACG - \triangle BIC - \triangle ABH = (1-3*\frac{3}{13})\triangle ABC = \frac{4}{13}*100 = \frac{400}{13}$

Hence, we get the area of $\triangle HIG = \frac{400}{13}$ unit$^2$.

Answer 3

(Adapted from my proof of the side-trisecting version on AoPS)

First, invert the logic of the construction. Start with an arbitrary triangle $GHI$, and extend $GH$ to $A$ so that $\frac{HA}{GH}=\frac12$, $HI$ to $B$ so that $\frac{IB}{HI}=\frac12$, and $IG$ to $C$ so that $\frac{GC}{IG}=\frac12$. Then, extend $AG$ to meet $BC$ at $D$, $BH$ to meet $AC$ at $E$, and $HI$ to meet $AB$ at $F$. We wish to find the ratio of the area of triangle $GHI$ to that of $ABC$.

(Unlike the figure in the question or that in Anirban Niloy's answer, this figure is actually to scale)

From $GH=2\cdot HA$ and the common vertex $I$, the red triangle $GHI$ has twice the area of the yellow triangle $HAI$. Similarly, $GHI$ has twice the area of the other two yellow triangles $IBG$ and $GCH$.
From $GH=2\cdot HA$ and the common vertex $C$, the yellow triangle $GHC$ has twice the area of the green triangle $HAC$. Similarly, $HIA$ has twice the area of $IBA$ and $IGB$ has twice the area of $GCB$, so the area of each yellow triangle is twice that of each green triangle.

With three green triangles, three yellow triangles, and one red triangle, the total area is $3+3\cdot 2+1\cdot 4=13$ times the area of one green triangle. The area of the red triangle $GHI$ is $4$ times the area of a green triangle, for a ratio of $\frac{4}{13}$ of the large triangle $ABC$.

That's the area ratios. But we're not done; we still need to show that this is the same as the original configuration. The ratio $\frac{AE}{EC}$ is equal to the ratio of areas $\frac{ABE}{CBE}$, which is equal to the ratio of areas $\frac{ABH}{CBH}$ - one green and one yellow triangle over one green, two yellow, and one red triangle. That's $1+2=3$ times a green triangle in the numerator, and $1+2\cdot 2+4=9$ times a green triangle in the denominator, a ratio of $\frac13$. Similarly, $\frac{BF}{FA}=\frac13$ and $\frac{CD}{DB}=\frac13$. The sides are indeed cut in fourths, and it's the same configuration.

Applying this to the given total area of $100$, the central triangle's area is $\frac{4}{13}\cdot 100=\frac{400}{13}$.