Commentary on Proposition VI.7

Back to Proposition VI.7.

Todhunter points out, after Walker, that some more words are necessary to make the enunciation precise: “If two triangles have one angle equal to one angle, the sides about other angles proportional <so that the sides subtending the equal angles are homologous>….”

This proposition is the extension to similar triangles of the ambiguous case already mentioned as omitted by Euclid in relation to equality of triangles in all respects (cf. note following I. 26, Vol. I. p. 306). The enunciation of VI. 7 has suggested the ordinary method of enunciating the ambiguous case where equality and not similarity is in question. Cf. Todhunter’s note on I. 26.

Another possible way of presenting this proposition is given by Todhunter. The essential theorem to prove is:

If two triangles have two sides of the one proportional to two sides of the other, and the angles opposite to one pair of corresponding sides equal, the angles which are opposite to the other pair of corresponding sides shall either be equal or be together equal to two right angles.

For the angles included by the proportional sides must be either equal or unequal.

If they are equal, then, since the triangles have two angles of the one equal to two angles of the other, respectively, they are equiangular to one another.

We have therefore only to consider the case in which the angles included by the proportional sides are unequal.

The proof is, except at the end, like that of VI. 7.

Let the triangles ABC, DEF have the angle at A equal to the angle at D;

let AB be to BC as DE to EF,

but let the angle ABC be not equal to the angle DEF.

A B C G D E F

The angles ACB, DFE shall be together equal to two right angles.

For one of the angles ABC, DEF must be the greater.

Let ABC be the greater; and make the angle ABG equal to the angle DEF.

Then we prove, as in VI. 7, that the triangles ABG, DEF are equiangular, whence

AB is to BG as DE to EF.

But AB is to BC as DE to EF, by hypothesis.

Therefore BG is equal to BC, and the angle BGC is equal to the angle BCA.

Now, since the triangles ABG, DEF are equiangular, the angle BGA is equal to the angle EFD.

Add to them respectively the equal angles BGC, BCA; therefore the angles BCA, EFD are together equal to the angles BGA, BGC, i.e. to two right angles.
It follows therefore that the angles BCA, EFD must be either equal or supplementary.

But (1), if each of them is less than a right angle, they cannot be supplementary, and they must therefore be equal;

  1. if each of them is greater than a right angle, they cannot be supplementary and must therefore be equal;

  2. if one of them is a right angle, they are supplementary and also equal.

Simson distinguishes the last case (3) in his enunciation: “then, if each of the remaining angles be either less or not less than a right angle, or if one of them be a right angle….”

The change is right, on the principle of restricting the conditions to the minimum necessary to enable the conclusion to be inferred. Simson adds a separate proof of the case in which one of the remaining angles is a right angle.

“Lastly, let one of the angles at C, F, viz. the angle at C, be a right angle; in this case likewise the triangle ABC is equiangular to the triangle DEF.

A B G C A B C G D E F

For, if they be not equiangular, make, at the point B of the straight line AB, the angle ABG equal to the angle DEF; then it may be proved, as in the first case, that BG is equal to BC.

But the angle BCG is a right angle;

therefore the angle BGC is also a right angle;
whence two of the angles of the triangle BGC are together not less than two right angles: which is impossible.
Therefore the triangle ABC is equiangular to the triangle DEF.”