Find out the area of a triangle with midpoints and quadrisection points

Question

In the above $\triangle$ABC, G is the midpoint of AC, D, E and F are quadrisection points of BC, AD and BG intersect at M, AF and BG intersect at N. The area of $\triangle$ABM is $7.2cm^2$ bigger than the area of quadrilateral FCGN. What is the area of $\triangle$ABC?

Please provide simple and elegant answers.

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I know that $\triangle$ABD and $\triangle$AFC have the same size, and $\triangle$ADF is twice as big as them. I also know that $\triangle$BAG and $\triangle$BCG have the same size. I tried building up a system of linear equations using the area of $\triangle$ABM, $\triangle$MBD, $\triangle$ANG and $\triangle$AFC as variables but failed.

Answer 1

We just need to find the ratios $DM : MA$ and $FN:NA$. They can both be obtained using Menelaus' Theorem. We have:

$$\frac {AG}{GC} \cdot \frac {CB}{BD} \cdot \frac {DM}{MA} = 1 \leadsto \frac {DM}{MA} = \frac14$$

$$\frac {AG}{GC} \cdot \frac {CB}{BF}\cdot \frac {FN}{NA} = 1\leadsto \frac {FN}{NA} = \frac34$$

Now we see that the area of $\triangle ABM$ is $\dfrac3{16}$ that of $\triangle ABC$, and the area of quadrilateral $FCGN$ is the sum of the areas of $\triangle CFN$ and $\triangle CNG$, i.e., $\dfrac 3{28} + \dfrac 2{28}$ that of $\triangle ABC$.

Finally we solve for the area of $\triangle ABC$ using the $7.5cm^2$ condition.

Answer 2

Applying Menelaus Theorem on $\triangle BGC$ considering $AMD$ will give $\frac {BM}{BG}=\frac {2}{5}$. $\Rightarrow Ar(\triangle ABM)=\frac {1}{5} \triangle ABC$

Now, applying Menelaus Theorem on $\triangle BGC$ considering $ANF$ as transversal will give $\frac {NG}{BG}=\frac {1}{7}$ $\Rightarrow Ar(\triangle ANG)=\frac {1}{14}\triangle ABC$

Now, $Ar(FCGN)$

$=Ar(\triangle AFC)-Ar(\triangle ANG)$

$=\frac {1}{4} \triangle ABC-\frac {1}{14} \triangle ABC$

$=\frac {5}{28} \triangle ABC$

Now put these values in the equation and solve for the area of $\triangle ABC$.

Answer 3

Here is an elementary method just building the linear equations as you mentioned, though it is of course not the most efficient.

say, area of $\triangle ABC = a$, $AN = m \times AF$ and $AM = n \times AD$.

Then you have,

$FCGN = \triangle ACF - \triangle AGN = \frac{a}{4} - \frac{a m}{8}$

$\triangle BNF = \frac{3a}{4} (1-m) = \frac{a}{2} - FCGN = \frac{a}{4} + \frac{am}{8} \implies m = \frac{4}{7}$.

Similarly, $\triangle ABM = \frac{an}{4}$

$\triangle AMG = \frac{n}{2} \times \frac{3a}{4} = \frac{3an}{8}$

$\triangle ABM + \triangle AMG = \frac{5an}{8} = \frac{a}{2} \implies n = \frac{4}{5}$

So, $\frac{a}{5} - \frac{5a}{28} = 7.2 \implies a = 336$