How do I find the area of a triangle formed by cevians? Given $\triangle ABC$, locate points $A_1$, $B_1$, $C_1$ on respective sides $BC$, $CA$, $AB$ such that
$$\frac{BA_1}{A_1C} =\frac{CB_1}{B_1A} = \frac{AC_1}{C_1B} = 2$$

How can I show that the area of the triangle formed by the intersections of $AA_1$, $BB_1$, and $CC_1$ is $1/7$ of the area of $\triangle ABC$?

 A: Observe how this whole problem is invariant under affine transformation. And any triangle can be mapped to any other via affine transformation. So if you have demonstrated the claim for one triangle, it must be true for all. Pick a particularly simple one. The numbers $3$ and $7$ seem to play a major role, so I chose two edge lengths to be $21$ and it works out nicely: all points are at integer coordinates, making the computation really easy:

The corners of the inner triangle are at
$$A_2=(12,3)\qquad B_2=(6,12)\qquad C_2=(3,6)$$
so its area is
$$\frac12\begin{vmatrix}12&6&3\\3&12&6\\1&1&1\end{vmatrix}=\frac{63}{2}$$
Compare that to the area of the whole triangle, and you get
$$\frac{63}{21^2}=\frac{63}{441}=\frac17$$
A: 
This layman proof depends on two relatively simple things:


*

*Area of a triangle = Half the base times its perpendicular height.
(An elementary school idea - but this also means that triangles on the same base with different heights, or vice-versa, have areas in proportion to their heights or bases. This can be handy sometimes.)

*Parallel projection of points on one line onto another line preserves the proportion of distances between adjacent points. That is, if $ a, b, c $ are points on one line $ L_1 $ which are projected via parallel lines to $a_1, b_1, c_1 $ onto another line $ L_2 $ then 
$$ |ab|:|bc| = |a_1b_1|:|b_1c_1| $$ (This is about US school grade 8 level transformation geometry.)
Proof:
Take any of the cevian triangles, e.g. $\Delta AEC$.
As its base |EC| is one third of the full side of $|BC|$, then its area must also be one third that of $\Delta ABC$.
Likewise for the other two cevian triangles, $\Delta LBC $ and $\Delta ABI $.
So we now have:
$ Area \Delta AEC = Area \Delta LBC = Area \Delta ABI  = \frac{1}{3} Area \Delta ABC $ ................. (1) 
From the diagram we see that each of the cevians $ \Delta AEC, \Delta MBC, \Delta ABI $ overlaps the other two at $ \Delta OEC, \Delta LBM, \Delta ANI $. So when we add the areas of the cevians together and subtract them from the area of the whole $ \Delta ABC $ to get the inner $ \Delta NMO $, we must realize that we are adding the areas of $ \Delta OEC, \Delta LBM $ and $ \Delta ANI $ twice unless we subtract them. 
So 
$ Area NMO = Area ABC - (Area AEC + Area LBC + Area ABI - Area OEC - Area LBM - Area ANI)   $
Applying relation (1) we get:
$  Area NMO = Area OEC + Area LBM + Area ANI $ ...............(2)  
So to get the area of the inner triangle $ \Delta NMO $ we must find the areas of the small corner triangles. Let's take $ \Delta OEC $ to start with.
Points I and J are $ \frac{1}{3} $ and $ \frac{2}{3} $ along $ |AC| $. We can see that a line parallel to BI that passes through point J will intersect |BC| at its mid-point G since J is the mid-point of |IC| and parallel project preserves line segment ratio. Point J is mapped onto point O on the intersection of cevian lines AE and CL. 
Since $ |AI| = |IJ| = |JC|  =>  |AN| = |NO| $ 
This same parallel projection also maps segment EON onto EGB. Since this results in a segment ratio of 
$$|EG|:|GB| = (\frac{1}{2} - \frac{1}{3}):\frac{1}{2} = 3:1 $$
we can conclude that $|ON| = 3 |EO| $
Since $|ON| = |NA| $ this means $|OA| = 6 |EO|$  implying $ |EA| = 7 |EO|  $
Projecting parallel to |AB|: 
Line segment EOA falls onto EPB which results in O being mapped to a point P such that 
$$ |EP| = \frac{1}{7} |EB| = \frac{1}{7} \frac{2}{3} |BC| = \frac{2}{21} |BC| $$ 
Thus 
$$ |CP| = (\frac{1}{3} + \frac{2}{21}) |BC| = \frac{3}{7} |BC| $$
and 
$$ |OP| = \frac{3}{7} |LB|  $$
This is all the information we need to estimate the area of $ \Delta OEC $ which has the same height as the $ \Delta OPC $, the latter being a similar triangle to $ \Delta LBC $.
$$ Area OPC = \frac{|CP|}{|BC|} \frac{|OP|}{|LB|} Area LBC $$
$$      = \frac{3}{7} \frac{3}{7} \frac{1}{3} Area ABC  = \frac{3}{49} Area ABC $$
Since $\Delta OEC $ has the same height as $ \Delta OPC $, we can estimate its area via the ratio of the triangle bases:
$$ Area OEC = \frac{|CE|}{|CP|} Area OPC = \frac{\frac{|BC|}{3}}{\frac{3|BC|}{7}} Area OPC  = \frac{7}{9} \frac{3}{49} Area ABC  $$
$$ = \frac{1}{21} Area ABC  $$
Similar reasoning gives us 
$$ Area LBM = Area ANI = Area OEC = \frac{1}{21} Area ABC  $$
Hence:
$$ Area NMO = Area OEC + Area LBM + Area ANI = \frac{1}{7} Area ABC  $$
This is the sort of insight that 'New Math' geometry can give you!
I was hoping that I could prove it simply by translating & folding the small triangles in the corners into the middle one. But the problem didn't oblige in this case due to overlaps. But Alex Chik has found a lovely graphical proof.
Thanks to Geobra for their free online sketching tool.
A: These are all excellent answers and I just want to add the general formula to the mix. There is a theorem known as Routh's theorem, which states that:

In geometry, Routh's theorem determines the ratio of areas between a given triangle and a triangle formed by the pairwise intersections of three cevians. The theorem states that if in triangle ${\displaystyle ABC}$ points ${\displaystyle D}$, ${\displaystyle E}$, and ${\displaystyle F}$ lie on segments ${\displaystyle BC}$, ${\displaystyle CA}$, and ${\displaystyle AB}$, then writing ${\displaystyle {\tfrac {CD}{BD}}}{\displaystyle =x}$, ${\displaystyle {\tfrac {AE}{CE}}}{\displaystyle =y}$, and ${\displaystyle {\tfrac {BF}{AF}}}{\displaystyle =z}$, the signed area of the triangle formed by the cevians $AD$, $BE$, and $CF$ is the area of triangle $ ABC$ times
$$\dfrac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)}$$
Plug $2$ into the above equation and you get the required answer immediately.
$\textbf{Hint}$: Use Menelaus Theorem to prove the above result.
