Last week I provided a puzzle involving equilateral triangles. You can find the puzzle details here
To explain the first solution, examine the drawing in Figure 1.
Figure 1: Solution 1
The altitude of the triangle ABC is h1 and the altitude of the triangle CEF is h2. G is the point where h1 intersect CA, and H is the point where h2 intersect the same line. It is fairly easy to prove that H is the same point as A, but not really necessary for our purposes. The triangles GBC and HEC are similar since they have two corresponding angles that are equal (both have a straight angle and share another angle). CE is twice the length of CB, therefore HE (h2) is twice the length of GB (h1). An area of a triangle is: ½bh (half base times altitude). Since the bases FC and CH of the triangles CEF and ABC are equal, but the altitude h2 is twice the length of the altitude h1, the area of CEF is twice the area of ABC. In other words, the area of CEF (as well as DEB and DAF) is 2S. Therefore the area of the triangle DEF is 3 × 2S + S = 7S.
The second solution is more creative. Examine the drawing in Figure 2.
Figure 2: Solution 2
You draw the lines EG and GF parallel to CF and EC, respectively, to form the parallelogram CEGF. Next, draw the lines BG, BH and CE. We know that FC = CB = BE = EG = GH = HF = HB = a. Triangles ABC, CHF and BGH are congruent since they all have two sides that are equal, plus the angle between them. This means that HC = BG = a. This means that the four triangles BEG, BGH, CBH and CHF enclosed by the parallelogram and ABC are congruent, therefore the area of the parallelogram is 4S. The triangle CEF has exactly half the area of the parallelogram, therefore the triangle’s area is 2S. Therefore the area of the triangle DEF is 3 × 2S + S = 7S.