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13 P. To find a mean proportional between two given straight lines AB, BC.
Place AB, BC in a straight line AC; bisect it (a); and, upon the bisectional point, describe a semicircle to pass through A, C; through B draw BD at right angles to AC (6); join AD, CD.
Because the angle ADC, in a semicircle, is a right angle (c); and that, in the right angled triangle ADC, a per pendicular BD 'is drawn from the right angle to the base; DB is a mean proportional between the segments of the base, which are the given straight lines AB, BC (d):
Wherefore, BD is the mean proportional sought.
(d) Cor. p. 8, 6.
14 Th. Equal parallelograms, which have equal an. gles, two and two, have their sides about equal angles reciprocally proportional: and parallelograms are equal, which have equal angles, two and two, and their sides :about equal angles reciprocally proportional.
Given two equal parallelograms AB, BC, hav. 1 ing equal angles at B. Place the parallelograms so that the angles at B shall be vertical to each other (a), and reciprocal sides, DB, BE and GB, BF, in straight lines: complete the parallelogram FE. Then DB : BE :: GB: BF; and if so, AB equals BC.
1. Because the parallelograms AB, BC are equal; and FE is another parallelogram ; AB: FE:: BC:FE (6). But AB:FE:: DB: BE, and BC: FE:: GB : BF (c); therefore, DB : BE :: GB: BF (d); and these proportionals are reciprocal sides about equal angles.
2. Because DB : BE :: GB : BF; and that DB : BE :: AB : FE, and GB: BF :: BC: FE (c); therefore AB : FE:: BC: FE (2); and so, the parallelograms AB and BC having the same ratio to the parallelogram FE, are equal to each other (e). Therefore, equal parallelograms, &c.
Q. E. D. Recite (a) p. 15, 1; (b) p. 7,5; (c) p. 1, 6;
(d) p. 11, 5; (e) p. 9, 5. Note.—The opposite sides and angles of a parallelogram are equal: p. 34 of b. 1.
15 Th. Equal triangles which have an angle of one equal to an angle of the other, have their sides about the equal angles reciprocally proportional: and triangles are
equal, which have an angle of one equal to an angle of the other, and the sides about the equal angles reciprocally proportional.
Given two equal triangles ABC, ADE, having equal angles at A. Place the triangles so that the angles at A may be vertical (a) to each other, and reciprocal sides CA, AD and EA, AB in straight lines : join BD. Then 1, CA:AD:: EA: AB; 2, ABČ=ADE.
1. Because the triangles ABC, ADE are equal; and that ABD is another triangle, "c ABC : ABD :: ADE : ABD (b): but ABC : ABD :: CA: AD; and ADE : ABD :: EA : AB (c); therefore CA: AD:: EA : AB (d); and, taken in this order, these proportionals are reciprocal sides.
2. Because CA : AD:: EA : AB; and that CA:AD:: ABC: ABD, and EA : AB :: ADE : ABD (c); therefore ABC:ABD:: ADE: ABD (6); and so, the triangles ABC, ADE, having the same ratio to ABD, are equal to each other (d). Therefore, equal triangles, &c.
Q. E. D. Recite (a)p. 14, 15, 1 ; def. 2, 6;
(b) p. 7,5; (c) p. 1, 6; (d) p. 11, 5; (e) p. 9,5. Note.-Reciprocal sides, def. 2, b. 6.
16 Th. If four straight lines be proportionals, the rectangles of the extremes and means are equal: and if the rectangles of the extremes and means be equal the four straight lines are proportionals.
1. Given AB to CD as E to F; sought ABXF=CDXE. 2. Given F XAB=EXCD; sought AB : CD::E:F.
From the points A, C, and at right angles to AB, CD, draw AG=F, and CH=E (a).
Complete the rectangles BG, DH.
1. Because AB : CD::E:F, and that E, F are equal to CH, AG; therefore e
H AB : CD :: CH : AG (6). Therefore the F sides of the parallelograms BG, DH, are a reciprocally proportional; and so, BG is equal to DH (c). Now, BG is the rectangle of AB and AG, or F; and DH is the rectangle of CD and EH, or E: therefore
А AB XF=CDXE.
2. Because F times AB equals E times CD; and that E, F are equal to CH, AG: therefore ABXAG=CDXCH. But parallelograms that are equal and equiangular, have their sides about equal angles reciprocally proportional (c): therefore AB : CD:: CH : AG; that is, AB : CD:: E:F. Therefore, if four straight lines, &c.
Q. E. D. Recite (a) p. 11, 1; (b) p. 7,5; (c) p. 14, 6.
17 Th. If three straight lines be proportionals, the rectangle of the extremes is equal to the square of the mean: and if the rectangle of the extremes be equal to the square of the mean, the three straight lines are proportionals.
Given three straight lines A, B, C;
B extremes (6): therefore AXC=BXD. But the rectangle of B and D is the square of B; because B=D. Therefore the rectangle of A, C equals the square of B.
But let the rectangle of A, C equal the square of B; then A:B:: B:C. For, if B=Ď, as before, then AXC=BXD: and if the rectangle of the extremes be equal to that of the means, the four straight lines are proportionals; therefore A:B::D:C (b). But B=D; therefore A is to B as B is to C. Wherefore, if three straight lines, &c.
Q. E. D. Recite (a) p. 7,5; (b) p. 16, 6.
18 P. Upon a given straight line (AB) to describe a rectilineal figure, similar and similarly situated to a given rectilineal figure (CDEF, or CDKEF).
Join DF; and make angles (a) at the points
Therefore, since the triangles ABG, CDF have
and BG : BH:: DF : DE (C): Therefore ex æquo AB : BH :: CD: DE(d); which are sides of the trapezia, about equal angles.
It may also be proved in like manner, that the other sides of the trapezia, about their equal angles, are proportionals; also, that the sides of the irregular pentagons, about their equal angles, are proportionals.
Therefore, the trapezium ABHG, and the irregular pentagon ABLHG, are severally similar
to the trapezium CDEF and pentagon CDKEF (e); and they are described upon the given straight line AB, as required.
And, in this manner, a figure of six or more sides, and similar to a given figure, may be described upon a given-straight line. Recite (a) p. 23, 1; (b) p. 32, 1; (c)p: 4, 6;
(d) p. 22, 5; (e) def. 1, 6 and def. 35, 1.
19 Th. Similar triangles are to one another in the duplicate ratio of their homologous sides.
Given the similar triangles ABC, DEF; having the angles at B, E, equal, and AB to BC as DE to EF: then BC, EF are homologous; because they are the consequents of the ratios (a); and the ratio of ABC to DEF is the duplicate or square o the ratio of BC to EF (6).
Take BG, a third proportional to BC, EF (c); so that BC: EF::EF: BG, and join GA. Then, since AB : BC :: DE : EF, alternately, AB : DE : : BC: EF (d). But BC : EF :: EF: BG; therefore AB : DE :: EF : BG(e): wherefore the triangles ABG, DEF are equal (f ).
And since BC, EF, BG are three proportionals, the ratio of BC to BG is the duplicate of the ratio of BC to EF (6): but BC: BG:: ABC : ABG (g), or its equal DEF (h); wherefore the ratio of ABC to DEF is the duplicate of the ratio of BC to EF. Therefore, similar triangles, &c.
Q. E. D. Recite (a) def. 10, 5; (6) def. 8, 5; (c) p. 11, 6;
(d) p. 16, 5; (e) p. 11, 5; (f) p. 15, 6;
(8) p. 1, 6; (h) p. 7,5. Cor. Hence, if three straight lines be proportionals, the first is to the third, as any triangle upon the first is to a similar triangle upon the second. The same is true of similar parallelograms, p. 41, 1.
20 Th. Similar polygons may be divided into the same number of similar triangles, having to each other the ratio of the polygons; which is the duplicate ratio of their homologous sides.
Let two polygons ABCDE, FGHKL, be similar, and have the sides AB, FG homologous (a).
Draw EB, EC and LG, LH; which divide the polygons into three triangles each.
Now these triangles are to be proved similar. 1. In the similar polygons (6),
H AB : AE :: FG: FL; and the an
gles are equal at A, F; so the triangles ABE, FGL being equal (c), have the angles ABE, FGL equal. But these equals are parts of the equal polygonal angles at B, G; therefore the other parts EBC, LGH are equal (d).
Also, in these equiangular trian
EB:BA:: LG: GF (e); And in the similar polygons AB : BC :: FG: GH(); Therefore, from equal distance EB : BC :: LG:GH ($).
Hence the triangles EBC, LGH, have their sides about equal angles proportional, and are therefore similar: for like reason, the triangles ECĪ, LHK are similar. Therefore, the similar polygons are cut into an equal number of similar triangles.
Again, the similar triangles have to each other the polygonal ratio ; namely, the duplicate ratio of AB to FG. For the ratio
of ABE to FGL is the duplicate of BE to GL (8)
of CDE to HKL is the duplicate of CE to HL. Therefore, the ratios of ABE to FGL, BCE to GHL, and CDE to HKL are all equal ; and, as one of the antecedents is to its consequent, so is the sum of all the antecedents to the sum of all the consequents (h). Thus the ratio of ABE to FGL equals that of ABCDE to FGHKL ; which is the duplicate ratio of the homologous sides AB, FG; and so of the other triangles.
Wherefore, similar polygons may be divided, &c. Q. E. D.
(d) ax. 3, 1; (e) p. 5, 6; (f) p. 22, 5;
21 Th. Rectilineal figures which are similar to the same rectilineal figures are also similar to one another
Let each of the figures A, B, be similar to C; A is similar to B.
For since A is similar to C, they are equiangular, and have their sides about equal angles, proportionals (a).
And since B is similar to C, they are equiangular, and have their sides about equal angles proportionals (a). Therefore the figures A, B, are, each
A of them, equiangular to C, and have the sides about equal angles of each of them, and of C proportionals. Wherefore, the rectilineal figures A and B are equiangular (b), and have their sides about equal angles proportionals (c). Therefore, A is similar to B
Q. E. D. Recite (a) def. 1, 6; (6) ax. 1, 1; (c) p. 11, 5.