THEOREM XXI, 551. In a right-angled triangle, any polygon upon the hypothenuse is equivalent to the sum of the similar and similarly described polygons on the other two sides. EXERCISES. 104. If 551 applies to semicircles, show the triangle equivalent to two crescent-shaped figures, called the lunes of Hippocrates of Chios (about 450 B.C.). PROBLEM IV. 552. To describe a polygon equivalent to one, and similar to another, given polygon. Let D be the one, and ABC the other, given polygon. By 262, on AC describe any CE = ABC, and on AE describe □ AM = D, and having ☀ AEM = × CAE. By 549, between AC and AF find a mean proportional GH. By 550, on GH construct the figure KGH similar and similarly described to the figure ABC. KGH is the figure required. It may be proved, as in 542, that AC and AF are in one line, and also LE and EM; therefore, by 505, But AC: AF :: CE : AM :: ABC : D. AC: AF: ABC : KGH, (547. If three sects form a proportion, a polygon on the first is to a similar polygon on the second as the first sect is to the third.) GEOMETRY OF THREE DIMENSIONS. BOOK VII. OF PLANES AND LINES. 553. Already, in 50, a plane has been defined as the surface generated by the motion of a line always passing through a fixed point while it slides along a fixed line. 554. Already, in 97, the theorem has been assumed, that, if two points of a line are in a plane, the whole line lies in that plane. Two other assumptions will now be made: 555. Any number of planes may be passed through any line. 556. A plane may be revolved on any line lying in it. 213 THEOREM I. 557. Through two intersecting lines, one plane, and only one, passes. HYPOTHESIS. Two lines AB, BC, meeting in B. CONCLUSION. One plane, and only one, passes through them. PROOF. By 555, let any plane FG be passed through AB, and, by 556, be revolved around on AB as an axis until it meets any point C of the line BC. The line BC then has two points in the plane FG; and therefore, by 554, the whole line BC is in this plane. Also, any plane containing AB and BC must coincide with FG. For let Q be any point in a plane containing AB and BC. Draw QMN in this plane to cut AB, BC, in M and N. Then, since M and N are points in the plane EF, therefore, by 554, Q is a point in the plane EF. Similarly, any point in a plane containing AB, BC, must lie in EF; therefore any plane containing AB, BC, must coincide with EF. 558. COROLLARY I. Two lines which intersect lie in one plane, and a plane is completely determined by the condition that it passes through two intersecting lines. 559. COROLLARY II. Any number of lines, each of which intersects all the others at different points, lie in the same plane; but a line may pass through the intersection of two others without being in their plane. 560. COROLLARY III. A line, and a point without that line, determine a plane. D B PROOF. Suppose AB the line, and C the point without AB. Draw the line CD to any point D in AB. Then one plane contains AB and CD, therefore one plane contains AB and C. Again, any plane containing AB must contain D; therefore, any plane containing AB and C must contain CD also. But there is only one plane that can contain AB and CD. Hence the plane is completely determined. 561. CorollarY IV. Three points not in the same line determine a plane. C B A For let A, B, C be three such points. Draw the line AB. Then a plane which contains A, B, and C must contain AB and C; and a plane which contains AB and C must contain A, B, and C. Now, AB and C are contained by one plane, and one only; therefore A, B, and C are contained by one plane, and one only. Hence the plane is completely determined. |