After a considerably long break, I resumed with two quite heavy posts on Gödel’s Incompleteness theorems. It’s time to crank down to more reasonable levels of mathematics. I recently came across the problem depicted with the figure below:

Point G is where the medians of the triangle ABC intersect. The triangle BDG is equilateral with side length l. Determine the lengths, AB, BC, and CA, of the sides of triangle ABC.

Since G is where the medians intersect, it follows that D, E, and F are the midpoints of BC, CA, and AB respectively.

This means that BD = DC = 1, giving us BC = 2.

Now, since triangle BDG is equilateral, it follows that ∠GDC = 120°. Also, since GD = DC, triangle GDC is isosceles, leading to the conclusion that ∠DGC = 30°.

Now, ∠BGD = 60° and ∠DGC = 30°. Hence, ∠BGC = 90°.

Now, consider triangle BGC. We have just obtained that ∠BGC = 90°. We also know that BG = 1 and BC = 2. Hence, using Pythagoras’ theorem, we can conclude that GC = √3.

Since G is the centroid of triangle ABC, G divides each of the medians in the ratio 2:1, with the large portion being between the centroid and the vertex.

Hence, GE = 1/2 and FG = √3/2.

Now, triangle EGC is a right angled triangle, right angled at G with GE = 1/2 and GC = √3. Once again using Pythagoras’ theorem, we get EC = √13/2, leading to CA = √13.

Similarly, triangle BGF is a right angled triangle, right angled at G with BG = 1 and FG = √3/2. Using Pythagoras’ theorem we get BF = √7/2, leading to AB = √7.

Of course, the key to solving this problem lay in realizing that BE and CF are perpendicular to each other, leading to a set of right angled triangles for each of which we can use Pythagoras’ theorem. If triangle BGD was not an equilateral triangle or if G was not the centroid of triangle ABC, the solution would not have been possible using pure geometry, which is what we have done. In that case, we would have needed to use trigonometry specifically as it relates to the angles and sides of a triangle. And in the next post we will move to studying the triangle properly.

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One response to “Primer to Properties of Triangles”

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