squares, = BKLH+KCML, the sum of the two paral lelograms 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.'

THEOREM XV.

Iu all circles the chord of 60 degrees is always equal in length to the radius.

fig. 38.

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. 1. theo. 5.) but since AC=CB, the angle CAB=CBA (by lemma preceding thico. 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 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 proportional, viz. fig. 34.

AB: ab:: AC: ac

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

fig. 34.

B

a

For the triangels being inscribed in two circles, it is plain since the angle A= a, the are BDC bdc, and consequently the chord BC is to b c, as the radius of the circle ABC is to be 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.

THEOREM XVII.

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,

A

fig. 35.
B

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 mul. tiplied by AB, will be equal to AE multiplied by AD. Q.,E, D.

THEOREM XVIII.

Triangles ABC, BCD, and parallelograms ABCF, and BDEC, having the same altitude, have the same proportion between themselves as their bases AB and BD. fig. 36.

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.

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 con. nected with the sign: the conclusion is always drawn from the first two and last two proportionals.

THEOREM XIX.

Triangles ABC. DEF, standing upon equal bases AB and DE, are to each other as their altitudes CG and FH. fig. 37.

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.

THEOREM XX.

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:: AE: ED. fig. 38.

fig. 38,
A

E

Draw CE and BD; the triangles BEC and EBD being on the same base BE and under the same paraltel 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.

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