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triangles a cd,dc b, by making the angle a cd= dcb (by postulate 4) then because a c=b c, and cd common, (by the last) the triangle a dc=d cb; and therefore i he angle a=b. 2. Ë. D.

Cor. Hence if from any point in a perpendicular which bisects a given line, there be drawn right lines to the extremities of the given one, they with it will form an isosceles triangle.

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THEO. VI.

Pl. l. fig. 25. The angle BCD at the centre of a circle ABED, is double the angle BAD at the circumference, standing upon the same arc BED.

Through the point A, and the centre C, draw the line ACĚ: then the angle ECD=CAD,+CDA; (by theo 4.) but since AC=CD being radii of the same circle, it is plain (by the preceding lemma) that the angles subtended by them will be also equal, and that their sum is double to either of them, that is, DAC+ ADC is double to CAD, and therefore ECD is double to CAD ; after the same manner BCE, is double to CAB, wherefore, BCE+ ECD, or BCD is double to BACHCAD or to BAD. Q. E. D.

Cor. 1. Hence an angle at the circumference is measured by half the arc it subtends or stands on.

Fig. 26.

Cor. 2. Hence all angles at the circumference of a circle which stands on the same chord as AB, are equal to each other, for they are all measured by half the are they stand on, viz. by half the arc AB.

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Cor. 3. Hence an angle in a segment greater than a semicircle is less than a right angle; thus ADB is measured by half the arc AB, but as the arc AB is less than a semicircle, therefore half the arc AB, or the angle ADB is less than half a semi

a circle, and consequently less than a right angle.

Fig. 27.

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Cor. 4. An angle in a segment less than a semicircle, is greater than a right angle, for since the arc AEC is greater than a semicircle, its half, which is the measure of the angle ABC, must be greater than half a semicircle, that is, greater than a right angle.

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Fig. 28.

Cor. 5. An angle in a semicircle is a right angle, for the measure of the angle ABD, is half of a semicircle AED, and therefore a right angle.

THEO. VII.

Pl. l. fig. 29.

If from the centre C of a circle ABE, there be let fall the fierpendicular CD on the chord AB. it will bisect it in the point D.

Let the lines AC and CB be drawn from the centre to the extremities of the chord, then since CA=CB, the angles CAB=CBA (by the lemma.) But the triangles ADC, BDC are right angled ones, since the line CD is a perpendicular ; and so the angle ACD=DCB ; (hy cor. 2. theo. 5.) then have we AC, CD, and the angle ACD in one triangle ; severally equal to CB, CD, and the angle BCD in the other : therefore (by theo. 6.) A=DB. 2. 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.

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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 DF is a perpendicular. Therefore (hy 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 &F=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 theo. 8.) therefore the corollary is plain.

THEO. X.

Pl. l.fig. 30.

In any triangle ABD, the half of each side is the sine of the opposite angle.

Let 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 chord of 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 same way may be proved that half of AD is the sine of ABD, and the half of AB tie sine of ADB. 2. E. D.

THEO. XI.

PL. 1.fig. 22.

If a right line GH cut two other right lines AB, CD, 80 as to make the alternate angles AEF, EFD equal to each other, then the lines AB and CD will be parallel.

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If it be denied that AB is parallel to CD, let IK be parallel to it; then IEF=(EFD)=AEF by par: 2. theo. 3.)a -greater to a less, which is absuid, whence IK is not parallel ; and the like we can prove of all other lines but AB; therefore AB is parallel to CD. 2. E. D.

THEO. XII.

PL. 1. fig. 3.

If two equal and parallel lines AB, CD, be joined by two other iincs AD, BC, those shall be also equal and parallel.

Let the diameter or diagonal BD be drawn, and we will have the triangles ABD, ('BD: whereof AB in on 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 angi- CBD=ADB, and thence the lines AD and BCate 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 paraliels 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.

THEO. XIII.

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 =

ČDF, (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=EFHG. 2. 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.

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