Carpets in Triangle
What Might This Be About?
1 March 2014, Created with GeoGebra
Let $M$ be a point in the interior $\Delta ABC.$ Three lines are drawn through $M$ parallel to the sides of the triangle, thereby producing three trapezoids. With a reference to the diagram below, in each of the trapezoids choose a diagonal so that the three have no common points: $AA',$ $BB',$ $CC'.$ These diagonals divide $\Delta ABC$ into seven pieces of which four are triangles.
Prove that the area of the central triangle equals the sum of the areas of the other three. More accurately,
$[XYZ] = [AB'Z] = [BC'X] + [CA'Y],$
where the brackets are used to denote the area of the included shape.
The three triangles $ABB',$ $BCC',$ $CAA'$ cover a certain area, that excludes $\Delta XYZ,$ but with triangles $AB'Z,$ $BC'X,$ $CA'Y$ covered twice.
Due to Euclid I.37, their total area is exactly $[ABC]:$
$[ABB'] + [BCC'] + [CAA'] = [ABM] + [BCM] + [CAM] = [ABC].$
Motivated by the Carpet Theorem, we observe that the area of their overlap equals the uncovered area.
[an error occurred while processing this directive]