In this proof of the puzzle discussed here we drop some perpendiculars, which is often a good start in Euclidean style geometry problems.
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Given a square ABCD, if a diagonal be drawn from B to D and another line drawn from A to the midpoint E of CD, so as to intersect the diagonal at F, the triangle formed by the two lines and the base AB of the square is one third of the square. Proof: We can take the side of the square to be 1, so its area will be 1. Construct perpendiculars EJ from E to BD, FH from F to AB, and FI from F to DA. Let the distance from A to H be d. Then the angles of AIFH are all right angles, so it is a rectangle, and side FI which is opposite to AH must also have length d. The point F lies on the diagonal BD, which bisects the angle at D, and therefore the perpendicular distances FI and FG are equal, and FG has length d also. Triangles AJI and AHF are similar, because they share a common angle FAH and both have a corresponding right angle. The distance JI is 1 and the distance AJ is .§ Therefore the ratio of EJ to AJ is 2:1 and that of FH to AH the same. The length of AH is d and there FH must be 2d. Then we have HG = 2d + d = 3d, and HG = 1 so FH = 2d = . The area of the shaded triangle AFB is base x height = x x 1 = , as claimed. |
§ We should probably prove this, which is easy enough but clutters uo the proof a bit.