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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, in as much as the triangle ABD=BDC, as now proved.

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Cor. 2. Hence also the triangle ADB on the same base AB, and between the same parallels with the pa. rallelogram 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.


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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.



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For AC=DB+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 trapezeum ABKC=KD

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the parallelogram ABCD=BDEF; in like manner we may prove the parallelogram EFGH=BDEF. Wherefore ABCD=BDEF=EFHG. 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 parallelograms.

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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 angle 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 (by the same ;) therefore half the square ABDE is equal to half the parallelogram BKLH, and the square ABDE equal to 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 legs; thus, the square root of the sum of the squares of the base and perpendicular, will be the hypothenuse.

Cor. 2. Having the hypothenuse 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.

fig. 33.

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 degrees, being measured by the arc AEB; therefore the sum of the other two angles is 120 degrees (by cor. theo. 5.) but since AC=CB, the angle CAB=CBA (by lemma preceding theo. 7.) consequently each of them will be

be equal to one another, as well 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.


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 legs opposite to the equal angles will be proportional, viz. fig. 34.

AB : ab :: AC : ac

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

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For the triangles being inscribed in two circles, it is plain since the angle Ara, the arc BDC=bdc, and consequently the chord BC is to b c, as the radius of the circle ABC is be 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 : ab : : 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 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.

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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 equi-angular 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.


Triangles ABC, BCD, and parallelograms ABCF, and BDEC, having the same altitude, have the same proportion between themselves as their bases AB and

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