Εικόνες σελίδας
PDF
[merged small][ocr errors]

If two sides of a triangle a b c be equal to each other, that is, ac = *b, the angles which are offiosite to those equal sides, will also be equal to each other ; viz. a = 6.

For let the triangle a b c be divided into two triangles a cd, d c b, by making the angle a c d = dc b (by postulate 4.) then because a c=b c, and c d common, (by the last) the triangle a doc=d c b : and therefore the angle a = b. Q. E. 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 isoceles triangle.

THEO. VII.

PL. 1. fig. 25.

The § BCD at the centre of a circle ABED, is double the angle BMD at the circumference, standing upon the same arc B.ED

Through the point A, and the centre C, draw the line ACE: 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 EUD is double to CAID, after the same manner BCE is double to CAB, wherefore, BCE + ECD, or BCD is double to HAC + CAD or to BAD. Q. E. D.

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

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 arc they stand on, viz. by half the arc A.B.

Fig. 26.

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 semicircle, and consequently less than a right angle.

Fig. 27.

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.

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.

[merged small][ocr errors]

If from the centre C of a circle ABE, there be let fall the fiershendicular CD on the chord AB, it will bisect it in the floint 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 ; (by cor. 2. theo. 5.) then have we AC,CD, and the angle ACD in one tri

* *

angle; severally equal to CB, CD, and the angle BCD in the other: therefore (by theo. 6.) AD = D.B. 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.

[ocr errors]

If from the centre of a circle ABE there be drawn a fiershendicular CD on the chord AB, and firoduced till it meets the circle in F, that line CF, will bisect the arc AB in the foint F.

Let the lines AF'and BF be drawn, then in the triangles ADF, BDF; AD= BD (by the last;) DFis common, and the angle ADF= BDF being both right, for CD or EF 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.) A F is half the arc, and AD half the chord A.B (by theo. 8.) therefore the corollary is plain.

THEO. X.
PL. 1...fig. 30.

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

[ocr errors]

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 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 same 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.

[ocr errors]

If a right line GH cut two other right lines AB, CD, so as to make the alternate angles MEF, EFD equal to each other, then the dines AB and CD will be farallel,

_ _ 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.

[ocr errors]

of two equal and farallel lines AB, CD, be joined by two other lines AD, BC, those shall be also equal and fiarallel.

Let the diameter or diagonal BD be drawn, and we will have the triangles ABD, CB D : whereof AB in one is=to CD in the other, B.D 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 # parallelogram, and the diagonal BD bisects the

o 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 paral-lelogram.

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. - r

THEO, XIII.
PL. 1...fig. 31.

...All harallelograms on the same or equal bases and between the same parallels, are equal to one another, that is, if BD = GH, and the lines ## .AF farallel, then the farallelogram. ABDC = BDFE= FHG.

For AC=BD=EF(by cor, the last;) to both add CE, then AE=CF, In the triangles A.B.E, 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 trapezium ABKC=F(DFE; to each of these add the triangle BKD, then the parallelogram ABCD=E DEF; in like manner we may prove the parallelogram EFGH-B DEF. 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.

[ocr errors]
« ΠροηγούμενηΣυνέχεια »