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Next, let the solid angle at A be contained by any number of plane angles BAC, CAD, DAE, EAF, FAB: these shall be together less than four right angles.

Let the planes in which the angles are, be cut by a plane, and let the common sections of it with those planes be BC, CD, DE, EF, FB.

Then, because the solid angle at B is contained by the three plane angles CBA, ABF, FBC, any two of them are together greater than the third, [XI. 20. therefore the angles CBA, ABF are together greater than the angle FBC.

For the same reason, at each of the points C, D, E, F, the two plane angles which are at the bases of the triangles having the common vertex A, are together greater than the third angle at the same point, which is one of the angles of the polygon BCDEF.

Therefore all the angles at the bases of the triangles are together greater than all the angles of the polygon.

Now all the angles of the triangles are together equal to twice as many right angles as there are triangles, that is, as there are sides in the polygon BCDEF; [I. 32. and all the angles of the polygon, together with four right angles, are also equal to twice as many right angles as there are sides in the polygon; [I. 32, Corollary 1. therefore all the angles of the triangles are equal to all the angles of the polygon, together with four right angles. [4x. 1.

But it has been shewn that all the angles at the bases of the triangles are together greater than all the angles of the polygon;

therefore the remaining angles of the triangles, namely, those at the vertex, which contain the solid angle at A, are together less than four right angles.

Wherefore, every solid angle &c. Q.E.D.

BOOK XII.

LEMMA.

If from the greater of two unequal magnitudes there be taken more than its half, and from the remainder more than its half, and so on, there shall at length remain a magnitude less than the smaller of the proposed magnitudes.

Let AB and C be two unequal magnitudes, 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, there shall at length remain a magnitude less than C.

D

A

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, and let DE be divided into DF, FG, GE, each equal to C.

From AB take BH, greater than its half, and from the remainder AH take HK greater than its half, and so on, until there be as many divisions in AB as in DE; and let the divisions in AB be AK, KH, HB, and the divisions in DE be DF, FG, GE.

[blocks in formation]

Then, because DE is greater than AB;

and that EG taken from DE is not greater than its half; but BH taken from AB is greater than its half;

therefore the remainder DG is greater than the remainder AH.

Again, because DG is greater than AH;

and that GF is not greater than the half of DG, but HK is greater than the half of AH;

therefore the remainder DF is greater than the remainder AK.

But DF is equal to C;

therefore C is greater than AK;

that is, AK is less than C. Q.E.D.

And 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, FGHKL be two circles, and in them the similar polygons ABCDE, FGHKL; and let BM, GN be the diameters of the circles: the polygon ABCDE shall be to the polygon FGHKL as the square on BM is to the square on GM.

B

M

Join AM, BE, FN, GL.
Then, because the polygons are similar,

therefore the angle BAE is equal to the angle GFL,
and BA is to AE as GF is to FL.

[VI. Definition 1. Therefore the triangle BAE is equiangular to the triangle GFL; [VI. 6.

therefore the angle AEB is equal to the angle FLG. But the angle AEB is equal to the angle AMB, and the angle FLG is equal to the angle FNG; therefore the angle AMB is equal to the angle FNG.

[III. 21.

And the angle BAM is equal to the angle GFN, for each of them is a right angle.

(III. 31.

A

B

M

Therefore the remaining angles in the triangles AMB, FNG are equal, and the triangles are equiangular to one another;

therefore BA is to BM as GF is to GN,

[VI. 4. [V. 16.

and, alternately, BA is to GF as BM is to GN; therefore the duplicate ratio of BA to GF is the same as the duplicate ratio of BM to GN. [V. Definition 10, V. 22. But the polygon ABCDE is to the polygon FGHKL in the duplicate ratio of BA to GF; [VI. 20. and the square on BM is to the square on GN in the duplicate ratio of BM to GN; [VI. 20. therefore the polygon ABCDE is to the polygon FGHKL as the square on BM is to the square on GN. [V. 11.

Wherefore, similar polygons &c. Q.E.D.

PROPOSITION 2.

THEOREM.

Circles are to one another as the squares on their diameters.

Let ABCD, EFGH be two circles, and BD, FH their diameters: the circle ABCD shall be to the circle EFGH as the square on BD is to the square on FH.

For, if not, the square on BD must be to the square on FH as the circle ABCD is to some space either less than the circle EFGH, or greater than it.

First, if possible, let it be as the circle ABCD is to a space S less than the circle EFGH.

[IV. 6.

In the circle EFGH inscribe the square EFGH.
This square shall be greater than half of the circle EFGH.

For the square EFGH is half of the square which can be formed by drawing straight lines to touch the circle at the points E, F, G, H;

and the square thus formed is greater than the circle; therefore the square EFGH is greater than half of the circle.

Bisect the arcs EF, FG, GH, HE at the points K, L, M, N ;

and join EK, KF, FL, LG, GM, MH, HN, NE. Then each of the triangles EKF, FLG, GMH, HNE shall be greater than half of the segment of the circle in which it stands.

For the triangle EKF is half of the parallelogram which can be formed by drawing a straight line to touch the circle at K, and parallel straight lines through E and F, and the parallelogram thus formed is greater than the segment FEK; therefore the triangle EKF is greater than half of the segment.

And similarly for the other triangles.

Therefore the sum of all these triangles is together greater than half of the sum of the segments of the circle in which they stand.

Again, bisect EK, KF, &c. and form triangles as before; then the sum of these triangles is greater than half of the sum of the segments of the circle in which they stand.

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