Angle-Side-Angle Theorem: Proposition 26

The color commentary in the Elements can be more interesting than the "propositions," or "proofs," themselves. And while the notes accompanying Proposition 26 are not nearly as quarrelsome as the Epicureans' comments about the Triangle Inequality Theorem, there's still some nice context here, in particular with regard to the idea that the theorem was used to determine the distance of ships from the coast.

The suggestions of Bretschneider and Cantor agree in the assumption that the necessary observations were probably made from the top of some tower or structure of known height, and that a right-angled triangle was used in which the tower was the perpendicular, and the line connecting the bottom of the tower and the ship was the base.

In Proposition 26, Euclid actually deals with two separate theorems: the Angle-Side-Angle Theorem (ASA) and the Angle-Angle-Side Theorem (AAS), but we'll only look at the first of these here.


Proof by Contradiction First, and Then the Ships

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Hover over the picture at the right to see the "Given"s in this proof: \(\small\measuredangle\mathtt{B} \cong \small\measuredangle\mathtt{E}\) and \(\small\measuredangle\mathtt {C} \cong \small\measuredangle\mathtt{F}\). Those are the two A's in ASA. And \(\overline{\small\mathtt{BC}}\cong\overline{\small\mathtt{EF}}\) is the S. Euclid says that he can show that the two triangles are congruent given this information.

So, he starts by supposing that he's wrong and that the triangles are not congruent. Suppose, he writes, that \(\overline{\small\mathtt{AB}}\) is actually longer than \(\overline {\small\mathtt{DE}}\):

Let \(\small\mathtt{AB}\) be greater, and let \(\small\mathtt{BG}\) [click on the triangles] be made equal to \(\small\mathtt{DE}\); and let \(\small\mathtt{GC}\) be joined.

Then, since \(\small\mathtt{BG}\) is equal to \(\small\mathtt{DE}\), and \(\small\mathtt{BC}\) to \(\small\mathtt{EF}\), the two sides \(\small\mathtt{GB}\), \(\small\mathtt{BC}\) are equal to the two sides \(\small\mathtt{DE}\), \(\small\mathtt{EF}\) respectively; and the angle \(\small\mathtt{GBC}\) is equal to the angle \(\small\mathtt{DEF}\); therefore the base \(\small\mathtt{GC}\) is equal to the base \(\small\mathtt{DF}\).

That last bit (that \(\small\mathtt{GC} = \small\mathtt{DF}\) because m\(\small\measuredangle\mathtt{GBC}\) = m\(\small\measuredangle\mathtt{DEF}\)) is Proposition 6, so it's cool. But we've also got that \(\small\Delta\)\(\small\mathtt{GBC}\) \(\cong\) \(\small\Delta\)\(\small\mathtt{DEF}\) without Proposition 6, because of the Side-Angle-Side Theorem. And if those two triangles are congruent, then \(\small\measuredangle\mathtt{GCB}\) is congruent to \(\small\measuredangle\mathtt{DFE}\). But that leads to a contradiction:

But the angle \(\small\mathtt{DFE}\) is by hypothesis equal to the angle \(\small\mathtt{BCA}\); therefore the angle \(\small\mathtt{BCG}\) is equal to the angle \(\small\mathtt{BCA}\), the less to the greater: which is impossible. Therefore \(\small\mathtt{AB}\) is not unequal to \(\small\mathtt{DE}\), and is therefore equal to it.

And if \(\small\mathtt{AB}\) is equal to \(\small\mathtt{DE}\), and all the "Given"s are true, then \(\small\Delta\)\(\small\mathtt{ABC}\) \(\cong\) \(\small\Delta\)\(\small\mathtt {DEF}\) by SAS.


So, About Those Ships

Apparently, we have a note telling us that the Angle-Side-Angle Theorem that you just proved was used to determine the distances of ships from the shore, but no indication is given as to how this was done. So, there has been some speculation:

Assuming that he was on top of a tower, he had only to use a rough instrument made of a straight stick and a cross-piece fastened to it so as to be capable of turning about the fastening (say a nail) so that it could form any angle with the stick and would remain where it was put. Then the natural thing would be to fix the stick upright (by means of a plumb-line) and direct the cross-piece towards the ship. Next, leaving the cross-piece at the angle so found, the stick could be turned round, still remaining vertical, until the cross-piece pointed to some visible object on the shore, when the object could be mentally noted and the distance from the bottom of the tower to it could be subsequently measured. This would, by [Proposition 26], give the distance from the bottom of the tower to the ship.