1. Since AB is parallel to CD, they may be considered as one broad line, crossed by another line, as GH; (then by the last theo.) GEB=CFH, and AEG-HFD.

2. Also GEB-AEF, and CFH-EFD; but GEB=CFH (by part 1. of this theo.) therefore AEF-EFD. The same way we prove FEB= EFC.

3. AEF=EFD; (by the last part of this theo.) but AEF=GEB (by theo 2.) Therefore GEB =EFD. The same way we prove AEG=CFE.

4. For since GEB = EFD, to both add FEB, then (by axiom 4.) GEB+FEB=EFD+FEB, but GEB + FEB, are equal to two right angles (by theo. 1.) Therefore EFD + FEB are equal to two right angles: after the same manner we prove that AEF+ CFE are equal to two right angles. Q. E. D.*

THEO. IV.

PL. 1. fig. 23.

In any triangle ABC, one of its legs, as BC, being produced towards D, it will make the external angle ACD equal to the two internal opposite angles taken together. Viz. to Band A.

Through C, let CE, be drawn parallel to AB; then since BD cuts the two parallel lines BA, CE; the angle ECD = B, (by part 3. of the last theo.) and again since AC cuts the same parallels, the angle ACE = A (by part. 2. of the last.) Therefore ECD + ACE = ACDB + A. Q. E. D.

Cor. 1. Hence, if a triangle have its exterior angle, and one of its opposite interior angles,

* For an excellent demonstration of this Theorem, (by the motion of the straight line crossing the parallel lines, about a point in one of them,) the read. er will consult Leslie's Geometry, Prop. 23, page 26.

double of those in another triangle; its remaining opposite interior angle will also be double of the corresponding angle in the other.*

That invaluable instrument, Hadley's Quadrant, is founded on this Corollary, annexed as an obvious consequence of the theorem. A ray of light SA, (Pl. 14. Fig. 2.) from the sun, against the mirror at A, is reflected at an angle equal to its incidence; and now striking the half silvered glass at C, it is again reflected to E, where the eye likewise receives, through the transparent part of that glass, a direct ray from the boundary of the horizon.

Hence, the triangle AEC has its exterior angle ECD, and one of its interior angles CAE, respectively double of the exterior angle BCD and the interior angle CAB, of the triangle ABC; wherefore the remaining interior angle AEC, or SEZ, is double of ABC; that is, the altitude of the sun above the horizon is double of the inclination of the two mirrors. But the glass at C remaining fixed, the mirror at A is attached to a moveable index, which marks their inclination.

The same instrument, in its most improved state, and fitted with a telescope, forms the sextant, which being admirably calculated for measuring angles in general, has rendered the most important services to Geography and Navigation.

THEO. V.

PL. 1. fig. 23.

In any triangle ABC, all the three angles, taken together, are equal to two right angles, viz. A + B + ĂCВ 2 right angles.

Produce CB to any distance, as D, then (by the

last) ACD=B÷A; to both add ACB; then ACD

This Corrollary, with the following demonstration is found in Leslies Geometry, pages 32 and 406.

+ACB= A + B + ACB; but ACD + ACB =2 right angles (by theo. 1.); therefore the three angles A+B+ ACB = 2 right angles. Q. E. D.

Cor. 1. Hence if one angle of a triangle be known, the sum of the other two is also known; for since the three angles of every triangle contain two right ones, or 180 degrees, therefore 180 -the given angle will be equal to the sum of the other two; or 180-the sum of two given angles gives the other one.

Cor. 2. In every right angled triangle, the two acute angles are=90 degrees, or to one right angle; therefore 90 one acute angle, gives the other.

THEO. VI.

PL. 1. fig. 24.

If in any two triangles, ABC, DEF, there be two sides AB, AC in the one, severally equal to DE, DF in the other, and the angle A contained between the two sides in the one, equal to D in the other; then the remaining angles of the one, will be severally equal to those of the other, viz. B E and C F: and the base of the one BC, will be equal to EF, that of the other.

If the triangle ABC be supposed to be laid on the triangle DEF, so as to make the points A and B coincide with D and E, which they will do, because AB DE (by the hypothesis); and since the angle A=D, the line AC will fall along DF, and inasmuch as they are supposed equal, C will fall in F; seeing therefore the three points of one coincide with those of the other triangle, they are manifestly equal to each other; therefore the angle B = E and C= F, and BC= EF. Q.E.D.

LEMMA.

PL. 1. fig. 11.

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

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

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

THEO. VII.

PL. 1. 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 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 CAD; after the same manner BCE is double to CAB, wherefore, BCE + ECD, or BCD is double to BAC+ CAD 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 arc they stand on, viz. by half the arc AB.

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.

THEO. VIII.

PL. 1. fig. 29.

If from the centre C of a circle ABE, there be let fall the perpendi cular 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; (by cor. 2. theo. 5.) then have we AC,CD, and the angle ACD in one tri