Let us go back to Terence Tao’s Solving Mathematical Problems, for another exercise. As done in a previous post, I would like to show an alternative solution to a problem the author proposes, that I think is instructive and considerably fast, once the appropriate perspective on the situation is taken.

Let $$ABFE$$ be a rectangle and $$D$$ be the intersection of the diagonals $$AF$$ and $$BE$$. A straight line through $$E$$ meets the extended line $$AB$$ at $$G$$ and the extended line $$FB$$ at $$C$$ so that $$DC \cong DG$$. Show that $\frac{\overline{AB}}{\overline{FC}} = \frac{\overline{FC}}{\overline{GA}} = \frac{\overline{GA}}{\overline{AE}}.\tag{1}\label{eq3702:1}$

Here is a picture that shows the initial situation.

Now, looking at \eqref{eq3702:1} as proportions, you can notice that the thesis is equivalent to stating that $$FC$$ is mean proportional between $$AB$$ and $$GA$$, and $$GA$$ is mean proportional between $$FC$$ and $$AE$$. This, in turn, should make you think of the Geometric Mean Theoremwhose demonstration would require the presence of some right-angled triangles; namely, a right-angled triangle with hypotenuse congruent to $$AB + GA \cong BG$$ and altitude congruent to $$FC$$, and a right-angled triangle with hypotenuse congruent to $$FC + AE$$ and altitude congruent to $$GA$$. Before reading further, try to find this configuration yourself starting from the Figure above (it should suffice to introduce just one additional point).

I proceeded as follows.

• Extend $$CD$$ and $$AE$$ so that they intersect at $$C’$$.
• Use the fact that $$BC \parallel AE$$ and ASA criterion on $$\triangle FDC$$ and $$\triangle C’DA$$ to show that $$C’A \cong FC$$. Conclude also that $$CD \cong C’D$$.

The situation now should be like the one shown below.

From what was stated above, the thesis is reached once we have proved that $$\triangle BC’G$$ and $$\triangle C’GE$$ are right-angled. Then, in fact, \eqref{eq3702:1} is a consequence of the Geometric Mean Theorem. How would you complete the proof? Read further if you need some hints.

Here is my approach.

• Use the hypotheses and the above demonstration to show that $$\triangle CC’G$$ is inscribed in a semicircle with center in $$D$$. Deduce that $$CG\perp C’G$$.
• Observe the quadrilateral $$BCEC’$$ and, noting that its diagonals bisect, prove that it is a parallelogram. Conclude that $$BC’ \parallel CG$$ and hence $$BC’ \perp C’G$$.