Commentary on Proposition III.21¶
Back to Proposition III.21.
Under the restriction that the “angle at the centre” used in III. 20 must be less than two right angles, Euclid’s proof of this proposition only applies to the case of a segment greater than a semicircle, and the case of a segment equal to or less than a semicircle has to be considered separately. The simplest proof, of many, seems to be that of Simson.
“But, if the segment BAED be not greater than a semicircle, let BAD, BED be angles in it: these also are equal to one another.
Draw AF to the centre, and produce it to C, and join CE.
Therefore the segment BADC is greater than a semicircle, and the angles in it BAC, BEC are equal, by the first case.
For the same reason, because CBED is greater than a semicircle,
Therefore the whole angle BAD is equal to the whole angle BED.”
We can prove, by means of reductio ad absurdum, the important converse of this proposition, namely that, if there be any two triangles on the same base and on the same side of it, and with equal vertical angles, the circle passing through the extremities of the base and the vertex of one triangle will pass through the vertex of the other triangle also. That a circle can be thus described about a triangle is clear from Euclid’s construction in III. 9, which shows how to draw a circle passing through any three points, though it is in IV. 5 only that we have the problem stated. Now, suppose a circle BAC drawn through the angular points of a triangle BAC, and let BDC be another triangle with the same base BC and on the same side of it, and having its vertical angle D equal to the angle A. Then shall the circle pass through D.
For, if it does not, it must pass through some point E on BD or on BD produced. If then EC be joined, the angle BEC is equal to the angle BAC, by III. 21, and therefore equal to the angle BDC. Therefore an exterior angle of a triangle is equal to the interior and opposite angle: which is impossible, by I. 16. Therefore D lies on the circle BAC.
Similarly for any other triangle on the base BC and with vertical angle equal to A. Thus, if any number of triangles be constructed on the same base and on the same side of it, with equal vertical angles, the vertices will all lie on the circumference of a segment of a circle.
A useful theorem derivable from III. 21 is given by Serenus (De sectione coni, Props. 52, 53).
If ADB be any segment of a circle, and C be such a point on the circumference that AC is equal to CB, and if there be described with C as centre and radius CA or CB the circle AHB, then, ADB being any other angle in the segment ACB, and BD being produced to meet the outer segment in E, the sum of AD, DB is equal to BE.
If BC be produced to meet the outer segment in F, and FA be joined,
Therefore the angle FAC is equal to the angle AFC. Also, by III. 21, the angles ACB, ADB are equal;
therefore their supplements, the angles ACF, ADE, are equal. Further, by III. 21, the angles AEB, AFB are equal. Hence in the triangles ACF, ADE two angles are respectively equal; therefore the third angles EAD, FAC are equal. But the angle FAC is equal to the angle AFC, and therefore equal to the angle AED. Therefore the angles AED, EAD are equal, or the triangle DEA is isosceles,
Adding BD to both, we see that
Now, BF being a diameter of the circle of which the outer segment is a part,
therefore AC, CB are together greater than AD, DB.
And, generally, of all triangles on the same base and on the same side of it which have equal vertical angles, the isosceles triangle is that which has the greatest perimeter, and of the others that has the lesser perimeter which is further from being isosceles.
The theorem of Serenus gives us the means of solving the following problem given in Todhunter’s Euclid, p. 324.
To find a point in the circumference of a given segment of a circle such that the straight lines which join the point to the extremities of the straight line on which the segment stands may be together equal to a given straight line (the length of which is of course subject to limits).
Let ACB in the above figure be the given segment. Find, by bisecting AB at right angles, a point C on it such that AC is equal to CB. Then with centre C and radius CA or CB describe the segment of a circle AHB on the same side of AB. Lastly, with A or B as centre and radius equal to the given straight line describe a circle. This circle will, if the given straight line be greater than AB and less than twice AC, meet the outer segment in two points, and if we join those points to the centre of the circle last drawn (whether A or B), the joining straight lines will cut the inner segment in points satisfying the given condition. If the given straight line be equal to twice AC, C is of course the required point. If the given straight line be greater than twice AC, there is no possible solution.