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Pl. 1. fig. 31 All parallelograms on the same or equal bases and between the same par. allels are equal to one another ; that is, if BD=GH, and the lines BH and AF are parallel, then the parallelogram ABDC=BDFE=EFHG.
For AC=BD=EF (by cor. the last); to both add CE, then AE=CF. In the triangles ABE, CDF, AB=CD and AE=CF, and the angle BAE=DCF (by part 3, theo. 3); therefore the triangle ABE=CDF (by theo. 6); let the triangle CKE be taken from both, and we will have the trapezium ABKC=KDFE; to each of these add the triangle BKD, then the parallelogram ABCD=BDEF: in like manner we may prove the parallelogram EFGH=BDEF. Wherefore ABDC=BDEF=EFGH. Q. E. D.
Cor. Hence it is plain that triangles on the same or equal bases and between the same parallels are equal, seeing (by cor. 2, theo. 12) they are the halves of their respective parallelogram.
Pl. 1. fig. 32. In every right-angled triangle, ABC, the square of the hypothenuse or longest side, BC, or BCMH, is equal to the sum of the squares made on the other two sides AB and AC, that is, ABDE and ACGF.
Through A draw AKL perpendicular to the hypothenuse BC, join AH, AM, DC, and BG; in the triangles BDC, ABH, BD=BA, being sides of the same square, and also BC=BH, and the included angles DBC=ABH (for DBA=CBH being both right, to both add ABC, then DBC=ABH), therefore the triangle DBC=ABH (by theo. 6); but the triangle DBC is half of the square ABDE (by cor. 2, theo. 12), and the triangle ABH is half the parallelogram BKLH. The same way it may be proved that the square ACGF is equal to the parallelogram KCLM. So ABDE+ACGF the sum of the squares =BKLH+KCML, the sum of the two parallelograms or square BCMH; therefore the sum of the squares on AB and AC is equal to the square on BC. Q. E. D.*
Cor. 1. Hence the hypothenuse of a right-angled triangle may be found by having the sides : thus, the square root of the sum of the squares of the base and perpendicular will be the hypothenuse.
* For different demonstrations of this excellent theorem, the reader may consult Leslie's Geometry (Prop. xi. book ii.).
Cor. 2. Having the hypothenuse and one side given to find the other; the square root of the difference of the squares of the hypothenuse and given side will be the required side.
Pl. 1. fig. 33. In all circles the chord of 60 degrees is always equal in length to the radius.
Thus in the circle AEBD, if the arc AEB be an arc of 60 degrees, and the chord AB be drawn, then AB=CB=AC.
In the triangle ABC the angle ACB is 60 degrees, being measured by the arc AEB; therefore the sum of the other two angles is 120 degrees (by cor. 1, theo. 5); but since AC= CB, the angle CAB=CBA (by lemma preceding theo. 7); consequently each of them will be 60, the half of 120 degrees, and the three angles will be equal to one another as well as the three sides : wherefore AB=BC=AC. Q. E. D.
Cor. Hence the radius from whence the lines on any scale are formed is the chord of 60 degrees on the line of chords.
Pl. 1. fig. 34. If in two triangles, ABC, abc, all the angles of one be each respectively equal to all the angles of the other ; that is, A=a, B=b, C=c; then the sides opposite to the equal angles will be proportional, viz.
AB: ab :: AC: ac
AB: ab : :BC: bc
and AC: ac :: BC: bc For the triangles being inscribed in two circles, it is plain, since the angle X=a, the arc BDC=* bdc and consequently the chord BC is to be as the radius of the circle ABC is to the radius of the circle abc (for the greater the radius is, the greater is the circle described by that radius; and consequently the greater any particular arc of that circle is, so the chord, sine, tangent, &c. of that arc will be also greater. Therefore, in general, the chord, sine, tangent, &c. of any arc is proportional to the radius of the circle); the same way the chord AB is to the chord ab in the same proportion. So AB: ab : :BC: bc. The same way the rest may be proved to be proportional.
* The arc BDC is not = in length to bdc, as might be supposed from the sign of equality ; but they contain the same number of degrees, as being the measure of equal angles.
Pl. l. fig. 35. If from a point A without a circle DBCE there be drawn two lines ADE, ABČ, each of them cutting the circle in two points, the product of one whole line into its external part, viz. AC into AB, will be equal to that of the other line into its external part, viz. AE into AD.
Let the lines DC, BE be drawn into two triangles ABE, ADC; the angle AEB=ACD (by cor. 2, theo. 7); the angle A is common, and (by cor. 1, theo. 5) the angle ADC=ABE; therefore the triangles ABE, ADC are mutually equiangular, and consequently (by the last) AC : AE :: AD: AB; wherefore AC multiplied by AB will be equal to AE multiplied by AD. Q. E. D.
Pl. 2. fig. 1.
Let any aliquot part of AB be taken which will also measure BD: suppose that to be Ag, which will be contained twice in AB, and three times in BD, the parts Ag, g B, Bh, hi, and iD being all equal, and let the lines gC, hC, and iC be drawn: then (by cor. to theo. 13) all the small triangles AgC, gCB, BCh, &c. will be equal to each other, and will be as many as the parts into which their bases were divided ; therefore it will be, as the sum of the parts in one base is to the sum of those in the other, so will be the sum of the small triangles in the first to the sum of the small triangles in the second triangle ; that is, AB:BD :: ABC: BDC.
Whence also the parallelograms ABCF and BDEC, being (by cor. 2, theo. 12) the
doubles of the triangles, are likewise as their bases. Q. E, D.
Note.--Wherever there are several quantities connected with the sign (: :) the conclusion is always drawn from the first two and last two proportionals.
PL. 2. fig. 26 Triangles ABC, DEF, standing upon equal bases AB and DE, are to each other as their altitudes CG and FH.
Let BI be perpendicular to AB and equal to CG, in which let KB=FH, and let AI and AK be drawn.
The triangle AIB=ACB (by cor. to theo. 13), and AKB= DEF; but (by theo. 18) BI : BK :: ABI: ABK. That is, CG: FH :: ABC : DEF. Q. E. D.
Pl. 2. fig. 3. If a right line BE be drawn parallel to one side of a triangle ACD, it will cut the two other sides proportionally, viz. AB : BC :: AỀ : ED.
Draw CE and BD; the triangles BEC and EBD being on the same base BE and under the same parallel CD, will be equal (by cor. to theo. 13) therefore (by theo. 18) AB : BC :: (BEA : BEC or BEA : BED::) AE: ED. Q. E. D.
Cor. 1. Hence also AC: AB :: AD: AE; for AC : AB (AEC : AEB : : ABD: AEB) :: AD: AE.
Cor. 2. It also appears that a right line which divides two sides of a triangle proportionally must be parallel to the remaining side.
Cor. 3. Hence, also, theo. 16 is manifest; since the sides of the triangles ABE, ACD, being equiangular, are proportional.
Pl. 2. fig. 4. If two triangles ABC, ADE have an angle BAC in the one equal to an angle DAE in the other, and the sides about the equal angles proportional ; that is, AB : AD :: AC: AE; then the triangles will be mutually equiangular.
In AB take Ad=AD, and let de be parallel to BC, meeting AC in e.
Because (by the first cor. to the foregoing theo.) AB: Ad (or AD) :: AC : Ae, and (by the hypothesis, or what is given in the theorem) AB : AD :: AC:AE; therefore Ae=AE, seeing AC bears the same proportion to each ; and (by theo. 6) the triangle Ade=ADE, therefore the angle Ade=D and Aed=E; but since ed and BCare parallel (by part 3, theo. 3) Ade=B, and Aed=C, therefore B=D and C=E. Q. E. D.
Pl. 2. fig. 5. Equiangular triangles ABC, DEF are to one another in a duplicate proportion of their homologous or like sides ; or as the squares AK and DM of their homologous sides.
Let the perpendiculars CG and FH be drawn, as well as the diagonals BI and EL.
The perpendiculars make the triangles ACG and DFH equiangular, and therefore similar (by theo. 16); for because the angle CAG=FDH, and the right angle AGC=DHF, the remaining angle ACG=DFH (by cor. 2. theo. 5). Therefore GC: FH ::(AC: DF::) AB:
DE, or, which is the same thing, GC: AB :: FH: DE, for FH multiplied by AB=GC multiplied by DE.
By theo. 19, ABC: ABI::(CG: AI or AB as before: : FH: DE, or DL): : DFE: ÒLE, therefore ABC: ABI:: DFE: DLE, or ABC: AK:; DFE: DM, for AK is double the triangle ABI, and DM double the triangle DEL, (by cor. 2. theo. 12.) Q. E. D.
Pl. 2. fig. 6.
Like polygons ABCDE, abcde, are in a duplicate proportion to that of the sides A B, ab, which are between equal angles A and B and a and b, or as the squares of the sides AB, ab.
Draw AD, AC, ad, ac.
By the hypothesis AB: ab:: BC: bc, and thereby also the angle B=b; therefore (by theo. 21) BAC=bac; and ACB =acb : in like manner EAD=ead, and EDA=eda. If therefore from the equal angles A and a, we take the equal ones EAD+BAC=ead+bac, the remaining angle DÂC=dac, and if from the equal angles D and d, EDA=eda be taken, we shall have ADC=adc: and in like manner if from C and e be taken BCA=bca, we shall have ACD=acd; and so the respective angles in every triangle will be equal to those in the other.
By theo. 22, ABC: abc : the square of AC to the square of ac, and also ADC: adc:: the square of AC to the square of ac; therefore, from equality of proportions, ABC: abc :: ADC : adc; in like manner we may show that ADC : adc :: EAD: ead. Therefore it will be, as one antecedent is to one .consequent, so are all the antecedents to all the consequents. That is, ABC is to abc as the sum of the three triangles in the first polygon is to the sum of those in the last. Or ABC will be to abc as polygon to polygon.
The proportion of ABC to abc (by the foregoing theo.) is as the square of AB is to the square of ab, but the proportion of polygon to polygon is as ABC to abc, as now shown: therefore the proportion of polygon to polygon is as the square of AB to the square of ab.