Notes
subdivided triangle ii solution

Solution to the Subdivided Triangle II Puzzle

Solution by Area of a Triangle and Dissection of Area

Consider the triangle with points labelled as above. From the formula for the area of a triangle, dividing the “base”, here $B C$, into three equal parts produces three subtriangles each of equal area and so each of area one third of the original. Therefore, triangles $A B D$, $A D E$, and $A E C$ all have area one third of the outer triangle.

Applying a similar argument to triangle $A E C$, but now with base $A C$ shows that triangles $E A F$ and $A F C$ both have area one half of $E A C$, and thus one sixth of the outer triangle.

Therefore, the orange region has area equal to one third plus one sixth of the outer triangle, which means that one half of the outer triangle is shaded.