See [url=https://www.geogebra.org/m/epzbrfxv]https://www.geogebra.org/m/epzbrfxv[/url] for diagram[br][br]AB means length(AB) and ABC means area(ABC)[br][br]WLOG assume ABC has area 1. It'll be natural anyway.[br][br]Elementary proof:[br][br]Since, [math]\frac{AA_3}{AB}=\frac{1}{3}[/math], [math]\frac{AA_3C}{ABC}=\frac{1}{3}[/math] and [math]\frac{A_3BC}{ABC}=\frac{2}{3}[/math]. By the same token [math]\frac{BB_3A}{ABC}=\frac{1}{3}[/math]. So [math]A_3BC-BB_3A=\frac{2}{3}-\frac{1}{3}\Longrightarrow B_3IC-A_3IA=\frac{1}{3}[/math].[br]Next, [math]A_3BC=BIC+A_3IB=\frac{2}{3}[/math][br]Recall that if two triangles have the same altitude their area is the ratio of their bases.[math]\frac{A_3A}{A_3B}=\frac{1}{2}\Longrightarrow\frac{A_3IA}{A_3IB}=\frac{1}{2}\Longrightarrow A_3IA=\frac{1}{2}A_3IB[/math][br][math]\frac{B_3C}{BC}=\frac{2}{3}\Longrightarrow\frac{B_3IC}{BIC}=\frac{2}{3}\Longrightarrow B_3IC=\frac{2}{3}BIC[/math][br]This yields two sets of linear equations[br][math]BIC+A_3IB=\frac{2}{3}[/math][br][math]\frac{2}{3}BIC-\frac{1}{2}A_3IB=\frac{1}{3}[/math][br]Multiplying the second one by 2 and equating them we get[math]BIC+A_3IB=\frac{4}{3}BIC-A_3IB\Longrightarrow2A_3IB=\frac{1}{3}BIC\Longrightarrow\frac{A_3IB}{BIC}=\frac{1}{6}\Longrightarrow\frac{A_3I}{IC}=\frac{1}{6}[/math][br]So since [math]\frac{A_3IA}{ICA}=\frac{1}{6}\Longrightarrow A_3IA=\frac{1}{6}ICA[/math] and [math]A_3AC=A_3IA+ICA[/math],[br][math]A_3CA=\left(1+\frac{1}{6}\right)ICA\Longrightarrow ICA=\frac{6}{7}A_3CA[/math][br]But [math]A_3CA=\frac{1}{3}[/math], so [math]ICA=\frac{6}{7}\cdot\frac{1}{3}=\frac{2}{7}[/math][br]The same method can be applied to [math]JAB[/math] and [math]KBC[/math], so they also have area [math]\frac{2}{7}[/math]. Now since [math]ABC=ICA+JAB+KBC+IJK[/math], [math]IJK=1-3\cdot\frac{2}{7}=\frac{1}{7}[/math]. So [math]\frac{IJK}{ABC}=\frac{1}{7}[/math][br][br]A faster proof using Menelaus' theroem, since it is more powerful:[br][br][math]\frac{AC_3}{C_3C}\cdot\frac{BC}{CB_3}\cdot\frac{B_3J}{JA}=1\Longrightarrow2\cdot3\cdot\frac{B_3J}{JA}=1\Longrightarrow\frac{B_3J}{JA}=\frac{1}{6}[/math][br][br]Which you would then use the last step of the above method to show that [math]JAB=\frac{2}{7}[/math] and then [math]IJK=\frac{1}{7}[/math][br]