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Plate I.

Let the diameter or diagonal BD be drawn, and we will have the two triangles ABD, CBD ; whereof AB in one is=to CD in the other, BD common to both, and the angle ABD=CDB; (by part 2. theo. 3.) therefore (by theo. 6.) AD=CB, and the angle CBD=ADB, and thence the lines AD and BC are parallel, by the preceding theorem.

Cor. 1. Hence the quadrilateral figure ABCD is a parallelogram, and the diagonal BD bisects the same, inasmuch as the triangle ABD=BCD, as now proved.

Cor. 2. Hence also the triangle ABD on the same base, AB, and between the same parallels with the parallelogram ABCD, is half the parallelogram.

Cor. 3. It is hence also plain, that the opposite sides of a parallelogram are equal ; for it has been proved that ABCD being a parallelogram, AB will be=CD and AD=BC,


All parallelograms on the same or equal bases and between the same parallels, are equal to one another, that is, if BD=GH, and the lines BH and AF parallel, then the parallelogram ABDC =BDFE=EFHG. fig. 31.

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 tlie triangle CKE be taken froin both, and we will have

Plate 1.

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.


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 (fig. 32.)

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

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Plate I.

squares =BELH + 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. Hence the hypothenuse of a right-angled triangle may be found by having the legs; thus, the square root of the sum of the squares of the base and perpendicular, will be the hypothenuse.

Cor. Having the hypothenuse and

and one leg given to find the other ; the square root of the difference of the squares of the hypothenuse and given leg, will be the required leg.


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. (Fig. 33.)

In the triangle ABC, the angle ACB is 60 de. grees, 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.

Plate I.

Cor. Hence the radius, from whence the lines on any scale are formed, is the chord of 60 degrees on the line of chords.


If in two triangles ABC, abc, all the angies of one be each respectively equal to all the angles of the other, that is, A=a, B=b, C=c: then the legs opposite to the equal angles will be proportional, viz.

Fig. 34.

AB : ab :: AC: ac

AB: ab :: BC: be and AC: ac :: BC: bc

For the triangles being inscribed in two circles, it is plain since the angle A=a, the arc BDC= bd c, and consequently the chord BC is to bc, as the radius of the circle ABC is to the radius of the circle a b c; (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 a b, in the same proportion. So AB: : a b :: BC: bc; the same way the rest may be proved to be proportional.


If from a point A without a circle DBCE there be drawn two lines ADE, ABC, each of them cutting the circle in two points ; the product of one

Plate I.

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. fig. 35.

Let the lines DC, BE, be drawn in the 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 equianguJar, 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.


Plate II. fig. 1.

Triangles ABC, BCD, and parallelograms

. ABCF and BDEC, having the same altitude, have the same proportion beiween themselves as their bases BA and BD.

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, gB, 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.

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