demonstration is long, we shall divide it into three distinct propositions. 1--Among triangles formed with two given sides, the greatest is that in which the two given sides make a

right angle— Let there be two triangles B A C and BA F80 D (fig. 80), having the side A B common, and A C=A

D. Then, if B A C is a right angle, we say the triangle BAC is greater than the triangle B A D, in which the augle at A is not a right angle. For, since the triangles have the same base A B they are to each other as their altitudes A C and D E. But A C is greater than DE, since its equal A D is greater than D E (30). 2- Among polygons in which all the sides but one are given, that is the greatest, of which all the given sides can be inscribed in a

semicircle having the unknown side for its diameter—. Let F 81 the polygon A B C D E F (fig. 81) be the greatest that

can be made of the given sides, A B, BC, CD, DE, E F, and the unknown side A F. Then we say that the angle formed by drawing lines from any vertex as D to the extremities of AF, is a right angle, and consequently inscribed in a semicircumference. For if A DF is not a right angle, then by the preceding proposition, the portion ADF may be enlarged without altering the portions A B C D and D E F; and thus the polygon itself would be enlarged. But by supposition it is already the greatest possible. Therefore the angle A D F is a right angle, and the same might be proved of all the other vertices. Consequently A F is the diameter of a semicircle in which the given sides are inscribed. 3-Among polygons formed of

given sides, the greatest is that which can be inscribed in a F 82 circle. Let the polygon A B C D E F G (fig. 82) be

inscribed; and let the polygon a b c d e f g, of which the sides are respectively equal to the former, be which cannot be inscribed. Then we say the formen is the greatest. Draw the diameter E M and join A M and M B. Then upon a b=A B make the triangle a bm=AB M, and join e m. Now, according to the preceding proposition, the polygon. E F G A M is greater than e f gam, unless this last can be inscribed in a semicircle having e m for its diameter, which by supposition cannot be done. For the same reason EDC BM is greater than e d c b . Hence the entire polygon E F G AM B C D E is greater than e f g amb c d e. Then subtracting the equal triangles A B M and a b m, we have the inscribed polygon A


B C D E F G greater than the polygon a b c d e f g which cannot be inscribed. We have now proved that the greatest polygon that can be formed with a given perimeter and a given number of sides, must be equilateral and capable of being inscribed in a circle. Then it must be regular; for it is equiangular as well as equilateral, since each of the inscribed angles has the same measure.

122. — Among regular polygons of the same perimeter but a different number of sides, that is the greatest which has the greatest number of sides ; and a circle is greater than any polygon of the same perimeter—. We shall need no diagram for this demonstration. We have already seen (104) that the area of a regular polygon is equal to its perimeter multiplied by half the radius of the inscribed circle. Consequently any two regular polygons are to each other as the products of their perimeters by half the radii of the inscribed circles. But in the case before us, the perimeters being the same, they are common factors. Therefore the two polygons are as the half radii or (66) as the radii of the inscribed circles. But the radii are as their circumferences (96). It only remains then to prove that of two circles inscribed in regular polygons of equal perimeters, that is the greater which is inscribed in the polygon of the greater number of sides. Now the inscribed circle is always less than the polygon unless the number of sides is infinite. This is self evident. It is equally evident, from mere inspection, that the difference becomes less as the number of sides of the polygon becomes greater. supposition we do not increase the perimeter by increasing the number of sides. Accordingly the limit remains always the same, and that circle must be the greatest in which the difference between it and the polygon is least. But this, as we have just seen, is when the number of sides is greatest, which was to be proved. It follows, moreover, that a circle is greater than any polygon of the same perimeter, because here the number of sides is infinite.

But by

Planes and their Angles.

123. We have bitherto considered planes as bounded by lines enclosing determinate areas. We are now to consider them as occupying certain relative positions with respect to each other. In this view, they are to be no longer limited in extent, but to be regarded as indefinitely pro"duced. And if we give them a determinate form in the diagrams, it is only because we cannot represent any other

to the eye.

124. Two straight lines meeting each other, determine the position of a plane“. By this we mean that a plane can have but one position in which the two straight lines

will lie in its surface. Let the two lines be A B and C F83 B (fig. 83). Let a plane be supposed to pass through A

B, that is to have A B in its surface. Then suppose this plane to turn about A B until the point C is in its surface. The line C B will be in its surface because two of its points are, and the position of the plane will be determined. For it is evident that it cannot be turned about either of the lines without departing from the other. Moreoverthree points not in the same straight line determine the position of a plane— For they may be joined by two straight lines, which we have just shown will determine the position of a plane.

125. The intersection of two planes is a straight lineIt is a line because the planes have no thickness. And it is a straight line, because, by the preceding proposition, if the points common to the two were not in the same straight line, the planes would have the same position, and consequently could not intersect each other.

126. Two planes which intersect each other, must form an opening of greater or less extent. This opening is called a plane angle. Thus the opening made by the planes

A B C D and A B E F which, for the sake of brevity, we F84 shall call A C and A E (fig. 84), is a plane angle. We

shall have a definite idea of this angle, if we suppose the plane A C at first to coincide with A E and then to turn about A B till it reaches its present position. The question now arises, how is this plane angle to be measured ? Suppose H I, in the plane A E, perpendicular to A B, and HG to have coincided with H I, while the planes coincided. Then during the motion of the plane A C, the point G would describe the arc G I. It is evident, moreover, that G I increases in the same proportion as the plane angle, since both are formed by one and the same motion. Therefore the arc G I may be taken for the measure of the plane angle. But the arc G I measures the linear angle

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G H I, formed by the two lines G H and I H perpendicular to A B. HenceThe angle made by two planes is measured by the angle made by two lines drawn perpendicular, at the same point, to the line of intersection. If then the angle G H I is acute, or less than 90°, the plane angle will be acute. If G H I is a right angle, the plane angle will be a right angle and the planes will be perpendicular. If GH I is obtuse, the plane angle will be obtuse. When the plane angle is either acute or obtuse, the two planes are said to be oblique or inclined to each other. Finally, since plane angles are measured by linear angles, it is evident that they have the properties of linear angles (20, 21, 22, 34, 35, 39).

127. -A line is said to be perpendicular to a plane when it is perpendicular to all the straight lines which can be drawn through its foot in the planem. For then it will make a right angle with the plane in every direction. This being admitted, we can demonstrate the following proposition-Å straight line is perpendicular to a plane when

it is perpendicular to two straight lines drawn through its foot in that plane— Let the plane be AC (fig. 85) and suppose E F 85 F perpendicular to F B and F C. We say that E F will be perpendicular to every other line F G drawn through its foot in the plane A C. Suppose F B to turn about E F remaining all the time perpendicular to it. Then one of its positions will be F C, since F C is by hypothesis perpendicular to EF. Now F B by its motion generates a plane to which E F is perpendicular, since it is perpendicular to the generating line in every position. This plane is B F C. But the plane B F C coincides with the plane AC (124), since by hypothesis the two planes have their position determined by the two lines F B and F C common to both. Hence tbe line F G is in the plane B F C. But E F is perpendicular to every line drawn through its foot in this plane. Therefore E F is perpendicular to F G. The same might be proved of any other line, since F G was taken at pleasure. Consequently E F is perpendicular to the plane A C, which was to be demonstrated.

128. -A perpendicular measures the shortest distance from a point to a plane-Two oblique lines drawn equally distant from the perpendicular are equal Of two oblique lines at unequal distances, the more distant is the greater- Let E F (fig. 86) be perpendicular to the plane A C, let E G and F 86

3. We say

E H be two oblique lines drawn equally distant from the
perpendicular, and let E I be drawn more distant than E
Gor E H. 1. We

that E F is shorter than


other "line. For E F is the side of a right triangle of which any other line as E G is the hypothenuse. 2. We say that É G=E H. For the triangle E F G=E F H, having the two sides, E F, F G, and the included angle E F G respectively equal to the two sides E F, FH and the included angle EFH. Therefore E G=E H. that E I is greater than E G or E H. E I and E H are drawn to the same straight line and E I is more remote than E H. Therefore (31) E I is greater than E H, and consequently greater than any other line as E G drawn at the same distance as E H.

129. Two planes are parallel when they are throughout at the same distance from one another. We have just proved that the shortest distance from a point to a plane is a perpendicular. Thereforetwo planes are parallel, or a straight line is parallel to a plane, when all the perpendiculars let fall from points in one to the other are equal— This being admitted, we can demonstrate the following proposition. -Two parallel lines comprehended between two paral

lel planes are equal— Let the two parallel planes be A B F87 and C D (fig. 87) If the two parallels are perpendicular

to the planes the proposition is evident from the defini. tion. But suppose they are not. Still we say they are equal. Let E F and G H be the two parallels oblique to the two planes. From E let fall the perpendicular È I to the plane C D, and from G let fall the perpendicular G K. Then the triangles FEI and HGK are equal. For E I=G K by definition, the angle I=the angle K being right angles, and the angle E=the angle G being complements of internal-external angles. Therefore (55) the two triangles are equal, and E F=G H.

130. The intersections of two parallel planes by a third F 88 are parallel lines, Let A B and C D (fig. 88) be two

parallel planes intersected by a third plane É F, and lét H G and E F be the intersections. Then we say that HG is parallel to EF. In the plane H F draw the parallel lines H E and GF. These are equal by the preceding proposition. Then join E G. The triangles H E G and E Ġ F are equal. For H E=G F, E G is common, and the angle H È G=the angle E G Í being alternate-inter

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