The one-seventh triangle

Reduction. An affine map multiplies all areas by the same constant and preserves ratios along lines. Assume $A=(0,0)$, $B=(1,0)$, $C=(0,1)$, area $\frac12$.

The points.
- $D=B+\frac23(C-B)=\left(\frac13,\frac23\right)$.
- $E=C+\frac23(A-C)=\left(0,\frac13\right)$.
- $F=\left(\frac23,0\right)$.

The cevians.
- $AD$: $y=2x$.
- $BE$: $y=\frac{1-x}3$.
- $CF$: $y=1-\frac{3x}2$.

Intersections.
- $AD\cap BE$: $2x=\frac{1-x}3\Rightarrow x=\frac17$: $P_1=\left(\frac17,\frac27\right)$.
- $BE\cap CF$: $\frac{1-x}3=1-\frac{3x}2\Rightarrow x=\frac47$: $P_2=\left(\frac47,\frac17\right)$.
- $CF\cap AD$: $2x=1-\frac{3x}2\Rightarrow x=\frac27$: $P_3=\left(\frac27,\frac47\right)$.

Area. With $\overrightarrow{P_1P_2}=\left(\frac37,-\frac17\right)$ and $\overrightarrow{P_1P_3}=\left(\frac17,\frac27\right)$:
$$[P_1P_2P_3]=\frac12\left|\frac37\cdot\frac27-\left(-\frac17\right)\cdot\frac17\right|=\frac12\cdot\frac7{49}=\frac1{14}.$$
Since $[ABC]=\frac12$, the ratio is $\frac{1/14}{1/2}=\frac17$. $\blacksquare$

Remark. This is a special case of Routh's theorem: for ratios $r$ in place of $2$, the area ratio is $\dfrac{(r-1)^2}{r^2+r+1}$; for $r=2$ it gives $\frac17$, for $r=1$ (medians) it gives $0$.

Lesson. Area-ratio problems often reduce, by affine invariance, to a computation in the coordinate triangle $(0,0),(1,0),(0,1)$.