Two hundred and ninety-fourth Meeting.

April 6, 1847.-MONTHLY MEETING.

The PRESIDENT in the chair.

Professor Strong, of New Brunswick, New Jersey, communicated the following papers, viz. :—

I. “An attempt to prove that the sum of the three angles of any rectilineal triangle is equal to two right angles.

"Def. Two quantities are said to be of the same kind, when the less can be multiplied by some positive integer, so as to exceed the greater. Thus, if A and B are quantities of the same kind, and if A is greater than B, then some positive integer, m, may be found, such that the inequality m B>A shall exist. For if m is taken greater than the quotient arising from the division of A by B, then evidently there results the inequality m B>A, as required.

"Dem. If m denotes any positive integer, then shall the inequality 2">m obtain. For the first member of the inequality denotes the product arising from taking 2 as a factor as often as these units in m, whereas the second member is the sum of the units represented by m, and the inequality is evident.

2m

"Cor. If we take A and B, as above, and m BA, there results B>; much more, then, shall the inequality B> have place. This follows at once since it has been shown that 2" is greater than m; and it is evident that if m is an indefinitely great number, definitely greater than 2

A

A

is in

"Ar. No angle of a rectilineal triangle can exceed two right angles. Prop. 1. To find a triangle that shall have the sum of its angles equal to the sum of the angles of any given triangle.

"Let A B C denote the given triangle; and suppose one of its sides,

BC, is bisected at D, and that D and the opposite angle A are connected by the right line A D, which is produced in the direction AD

to E, so that DEAD, and that the point E and the angle C are connected by a right line; then shall the sum of the angles of the triangle AC E equal that of the triangle A B C.

"For the opposite vertical angles AD B, C DE are equal (Simson's Euclid, B. I., prop. 15), and by construction BDD C, AD =DE; hence (Sim., B. I., p. 4) the triangles ADB, CDE are identical; so that their bases AB, CE are equal, and their angles A B D, E C D are equal; also the angle BAD equals the angle CED. Hence the angle C of the triangle A CE equals the sum of the two angles B and C of the given triangle (A B C), and the sum of the angles A and E of the triangle ACE is equal to the angle A of the given triangle (ABC); .. the sum of the angles of the triangle ACE is equal to that of the given triangle ABC, as required.

"Cor. 1. Let the angle B A C of the given triangle be denoted by A; then if CE (= A B) is not greater than A C, the angle CAE is not greater than CEA (Sim., props. 5, 19, B. I.); hence the angle A of the triangle ACE is not greater than ; we shall call the triangle ACE the first derived triangle. Of the two sides AC and CE of the triangle A CE, let CE be that which is not the greater; and let a right line be drawn from the angle A, opposite to the side CE, through the point, H, of bisection of C E, and suppose the line thus drawn to be produced in the direction AH to I, so that HI= HA; then connect the point I and the angle C of the triangle A C E by a right line; and there will be formed the triangle A CI. In the same way that it was shown that the sum of the angles of the triangle ACE is equal to the sum of the angles of the (given) triangle A B C, it may be shown that the sum of the angles of the triangle A CI is equal to that of ACE; consequently the sum of the angles of ACI equals that of the given triangle ABC. And if AC is not greater than CI, then it may be shown (as before) that the angle I is not greater than the angle CA E÷2, and since CA E is not greater than 4,.. the angle I is not greater than, we shall call A CI the second derived triangle.

"We may in the same way (that we derived the triangle A CI from ACE) derive a triangle from ACI (called the third derived triangle), having the sum of its angles equal to that of AC I, and of course equal to that of the given triangle A B C, and having one of its angles not greater than the angle AIC÷2, and consequently not greater

A

23.

than And proceeding in the same way from triangle to triangle, until we obtain the mth derived triangle, then the sum of its angles will equal that of the given triangle A B C, and one of its angles will not be greater than; where m is of course a positive integer.

Ꭺ

A

"Cor. 2. If we obtain the derived triangle whose number is m+1, the sum of two of its angles will equal that angle of the mth triangle which has been shown not to be greater than; we hence see how from any given triangle to derive another triangle such that the sum of its angles shall equal that of the given triangle, and such that the sum of two of its angles shall not be greater than where A denotes one 2 in ; of the angles of the given triangle.

A

"Remark. Cor. 1 is substantially the same as Mr. I. Ivory's process, given at page 189 of the New York edition of J. R. Young's Elements of Geometry.

"Prop. 2. The sum of the angles of any triangle is not greater than two right angles.

A

"Let the triangle A B C, of Prop. 1, represent any triangle, and denote a right angle by R, and if possible let the sum of the angles of the triangle equal 2 A+ V, V being a finite positive angle. Then, using A to represent the angle B A C of the triangle, some positive integer, m, may be found so that the inequality m V>A shall exist. From m VA, it follows that V>>, or V is greater than By Cor. 2, Prop. 1, we may derive a triangle from ABC, such that the sum of two of its angles shall not be greater than hence the sum of these two angles is less than V; consequently the third angle of the triangle must be greater than 2 R, which is impossible. Hence the sum of the angles of the triangle A B C is not greater than two

right angles.

2

A

"Prop. 3. The sum of the angles of any triangle is greater than a right angle.

as to make the angles DC F, GFH each equal to the angle BAC, and connect the points D and F, D and G, D and B, by right lines; also through D draw D K, at right angles to D G. Since A B DC, A C=C F, and the angle BAC equals the angle D C F, the triangles ABC, C D F are identical (Sim., B. I., p. 4), making the sides DF and BC equal to each other, and the angle CDF equal to the angle ABC, and the angle CFD equal to the angle ACB; hence the sum of the angles BA C, BCA equals the sum of the angles BCA, DCF, which equals the sum of the angles CFD, GF H. Since the sum of the three angles at C makes two right angles, and that the sum of the angles at F makes two right angles (Sim., B. I., p. 13), it follows from what has been proved that the angles B C D, DFG are equal, and since B C DF, CD- FG, the triangles BCD, DFG are identical (Sim., B. I., p. 4), making BD equal to D G, and the angle CBD equal to the angle FD G, and the angle CDB equal to the angle FGD; hence the sum of the angles CBD, CDB equals the sum of the angles C D B, FD G. By Prop. 2, since the sum of the three angles of any triangle is not greater than two right angles, and that the three angles at C make two right angles, it follows that the sum of the angles B and D of the triangle B C D is not greater than the sum of the angles ACB, FCD; hence, and from what has been proved, it follows that the sum of the angles of the triangle ABC is not less than the sum of the angles B D C, C D F, FDG; but it is evident that the sum of these angles exceeds the right angle K DG by a fine angle; hence the sum of the three angles of the triangle ABC exceeds a right angle, as required.

"Remark. If A B is extended in the direction A B to E, so that BE = BD, and if the points D and E, G and E, are connected by right lines, the point D falls evidently within the pentagonal figure CBEGF; and if R denotes a right angle, the sum of the angles B DC, C D F, FDG, G DE, EDB is equal to 4 R (Sim., B. I., p. 15, Cor. 2). Since the sum of the angles A B D, DBE is equal to 2 R, the sum of the angles at the base of the isosceles triangle B D E is not greater than the angle ABD; consequently the angle BDE is not greater than the angle A BD÷2, which is not greater than half the sum of the angles B D C, C D F, FDG. We now observe that the sum of the angles of the triangle ABC is not less than R+R. For if the sum of the angles of the triangle A B C is not greater than R+R, then by what has been shown the sum of the angles BDC,

CDF, FDG, BDE is not greater than R+R+ 1⁄2 + 1⁄2 R §R; consequently the angle D of the triangle EDG is not less than 4 RR = §§ R=2R+, which is impossible; hence the sum of the angles of any triangle, A B C, is greater than R+R, as required.

"Prop. 4. The sum of the angles of any triangle is not less than two right angles.

"If the sum of the angles of any one triangle is not the same as that of any other, then there are evidently some triangles having the sum of their angles less than that of any others; let, therefore, A BC (see fig., Prop. 1) denote a triangle such that the sum of its angles is not greater than that of any other triangle. If R represents a right angle, then, since the sum of the an. gles of any triangle is greater than R, the sum of the angles of the triangle ABC may be expressed by R+V, V being a positive angle. If we denote the angle BAC of the given triangle by A, then, as in Prop. 2, we may find some positive E integer, m, such that the inequality V> shall have place; and by v> 2m Cor. 1 and 2 of Prop. 1 we may derive from the triangle A B C another triangle represented by DEF such that the sum of its angles shall equal that of ABC, and consequently equal R+V, and further

A

A

KH

such that the sum of two of its angles, EDF, EFD, shall not be greater than ;.. the sum of these (two) angles is less than V, consequently the third angle E of the triangle is greater than R. Of the two sides DE, E F, let DE be that which is not the less, and through E draw E K, at right angles to DE; then, since the angle DEF is greater than R, the perpendicular will of course meet the side DF at some point, as K, between D and F. Since the sum of the sides DE and E F is greater than D F (Sim., p. 20, B. I.), and that DK is greater than DE (Sim., p. 19, B. I.), we of course have DK greater than KF. Hence extend D F to G, so that G K=DK, and extend EK to H, so that KHE K, and connect G and H