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angle ; severally equal to CB, CD, and the angle BČD in the other: therefore (by theo. 6.) AD =DB. Q. E, D.
Cor. Hence it follows, that any line bisecting a chord at right angles, is a diameter; for a line drawn from the centre perpendicular to a chord, bisects that chord at right angles ; therefore, conversely, a line bisecting a chord at right angles must pass through the centre, and consequently be a diameter.
Pl. 1. fig. 29.
If from the centre of a circle ABE there be drawn a perpendicular CD on the chord AB, and produced till it meets the circle in F, that line CF, will bisect the arc AB in the point F.
Let the lines AF and BF be drawn, then in the triangles ADF, BDF; AD=BD (by the last;) DF is common, and the angle ADF=BDF being both right, for CD or BF is a perpendicular. Therefore (by theo. 6.) AF= FB ; but in the same circle, equal lines are chords of equal arcs, since they measure them (by def. 19.): whence the arc AF=FB, and so AFB is bisected in F, by the line CF.
Cor. Hence the sine of an arc is half the chord of twice that arc.
For AD is the sine of the arc AF, (by def. 22.) AF is half the arc, and AD half the chord AB (by tbeo. 8.) therefore the corollary is plain.
PL. 1. fig. 30. In any triangle ABD, the half of each side is the sine of the opposite angle.
Lot the circle ADB be drawn through the points A, B, D,
then the angle DAB is measured by half the arc BKD, (by cor 1. theo. 7.) viz. the arc BK is the measure of the angle BAD; therefore (by cor, to the last) BE the half of BD is the sine of BAD: the saine way may be proved that half of AD is the sine of ABD, and the half of AB the sine of ADB. Q. E. D..
PL. 1. fig. 22
If a right line GH cut two other right lines AB, CD, 80 as to make the alternate angle: AEF, EFD equal to each other, then the lines AB and CD will be parallel,
If it be denied that AB is parallel to CD, let IK be parallel to it; then IEF=(EFD)=AEF (by part 2. theo. 3.) a greater to a less, which is absurd, whence IK is not parallel ; and the like we can prove of all other lines but AB; therefore AB is parallel to CD. Q. E. D.
Pl. 1. fig. 3,
If two equal and parallel lines AB, CD, be joined by two other kines AD, BC, those shall be also equal and parallel.
Let the diameter or diagonal BD be drawn, and we will have the 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 e 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 paralHelogram.
Cor. 3. Jt 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.
Pl. 1. fig. 31.
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.
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 ČKE 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 hypothepuse 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 D BC 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.
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.
* For different demonstrations of this excellent Theorem, the reader may consult Leslies Geometry, (Prop. xi. book ii.)
PL. 1. Ag. 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=CB A (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 anglee of one be each respect tively equal to all the angles of the other, that is, a, B=6, Cc: then the sides opposite to the equal angles will be propor tional, 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 A=a, the arc BDC=* b dc, and consequently the chord BC is to b C, as the radius of the circle ABC is to the radius of the circle a b c;(for the greater the radius is, the
* The arc BDC is not a in length to bdc, as might be supposed from the sigo of equality ; but they contain the same number of degrees, as being the do casure of equal angles,