are sides in the polygon BCDEF; and all the angles of the polygon, together with four right angles, are likewise equal to twice as many right angles as there are sides in the polygon (I. 32, Cor. 1); therefore 3. All the angles of the triangles are equal to all the angles of the polygon together with four right angles (I. Ax. 1); but all the angles at the bases of the triangles are greater than all the angles of the polygon, as has been proved; therefore 4. The remaining angles of the triangles-viz., those at the vertex, which contain the solid angle at A, are less than four right angles. Wherefore, the plane angles, &c. Q.E.D. BOOK XII. (PROPOSITIONS 1 AND 2.) LEMMA.* [This lemma, which is usually inserted here, is the first proposition of Euclid, Book X. It is necessary to the understanding of some of the propositions of Book XII.] If from the greater of two unequal magnitudes of the same kind, there be taken more than its half, and from the remainder more than its half; and so on: there will at length remain a magnitude less than the least of the proposed magnitudes. Let AB and C be two unequal magnitudes, of the same kind, of which AB is the greater. If from AB there be taken more than its half, and from the remainder more than its half, and so on; Then there will at length remain a magnitude less than C. D H B Construction. For C may be multiplied so as at length to become greater than AB. Let it be so multiplied, and let DE, its multiple, be greater than AB; also let DE be divided into parts DF, FG, and GE, each equal to C. From AB take BH greater than its half, and *A Lemma is a proposition of no importance in itself, merely introduced for the purpose of demonstrating some other proposition. from the remainder AH take HK greater than its half, and so on, until there be as many divisions in AB, as there are in DE; and let the divisions in AB be AK, KH, and HB; and the divisions in DE be DF, FG, and GE. Demonstration. Because DE is greater than AB, and EG taken from DE is not greater than its half, but BH taken from AB is greater than its half; therefore 1. The remainder GD is greater than the remainder HA. Again, because GD is greater than HA, and that GF is not greater than the half of GD, but HK is greater than the half of HA; therefore 2. The remainder FD is greater than the remainder AK; but FD is equal to C; therefore 3. C is greater than AK; that is, AK is less than C. Wherefore, if from the greater, &c. Q.E.D. Corollary. If only the halves be taken away, the same thing may in the same way be demonstrated. PROPOSITION 1.-Theorem. Similar polygons inscribed in circles, are to one another as the squares on their diameters. Let ABCDE and FGHKL be two circles having the similar polygons ABCDE and FGHKL inscribed in them; and let BM and GN be the diameters of the circles. Then the polygon ABCDE is to the polygon FGHKL as the square on BM is to the square on GN. Construction. Join BE, AM, GL, and FN. Demonstration. Because the polygon ABCDE is similar to the polygon FGHKL; therefore B E M L 1. The angle BAE is equal to the angle GFL; and BA is to AE as GF is to FL; therefore the two triangles BAE, GFL are equiangular (VI. 6); and therefore 2. The angle AEB is equal to the angle FLG. But the angle AEB is equal to the angle AMB, because they stand upon the same arc (III. 21). For the same reason, the angle FLG is equal to the angle FNG; therefore also 3. The angle AMB is equal to the angle FNG; And the angle BAM is equal to the angle GFN (III. 31); therefore 4. The remaining angles in the triangles ABM and FGN are equal, and they are equiangular to one another; wherefore BM is to GN as BA is to GF (VI. 4); therefore 5. The duplicate ratio of BM to GN is the same with the duplicate ratio of BA to GF (V. Def. 10 and V. 22) ; but the ratio of the square on BM to the square on GN is the duplicate ratio of that which BM has to GN (VI. 20); and the ratio of the polygon ABCDE to the polygon FGHKL is the duplicate of that which BA has to GF (VI. 20); therefore 6. As the polygon ABCDE is to the polygon FGHKL so is the square on BM to the square on GN. Wherefore, similar polygons, &c. Q.E.D. PROPOSITION 2.-Theorem. Circles are to one another as the squares on their diameters. Let ABCD and EFGH be two circles, and BD and FH their diameters. Then the square on BD is to the square on FH as the circle For, if not, the square on BD is to the square on FH as the circle ABCD is to some space either less than the circle EFGH, or greater than it. First, let it be to a space S less than the circle EFGH; and in the circle EFGH inscribe the square EFGH (IV. 6). This square is greater than half of the circle EFGH; because, if through the points E, F, G, and H, tangents be drawn to the circle, the square EFGH is half of the square described about the circle (I. 41). But the circle is less than the square described about it; therefore 1. The square EFGH is greater than half the circle. Bisect the arcs EF, FG, GH, and HE at the points K, L, M, and N, and join EK, KF, FL, LG. GM, MH, HN, and NE; then 2. Each of the triangles EKF and FLG is greater than half of the segment which contains it. For if straight lines touching the circle be drawn through the points K, L, M, and N, and the parallelograms upon the straight lines EF, FG, GH, and HE be completed, each of the triangles EKF, FLG, GMH, and HNE is the half of the parallelogram which contains it (I. 41); but every segment is less than the parallelogram which contains it; therefore 3. Each of the triangles EKF, FLG, GMH, and HNE is greater than half the segment which contains it. |