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THEOREM IV.

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 pposite angles taken together. Viz. to B and A. fig. 23.

fig. 23. E

Thro' 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=ACD B+A. Q. E. D.

THEOREM V.

In any triangle ABC, all the three angles taken together are equal to two right angles, viz, A+B+ACB =2 right angles. fig. 23.

Produce BC to any distance, as D, then (by the last.) ACD-B+A; to both add ACB; then ACD+ 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.

Cer. 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 acate angles are 90 degrees, or to one right-angle : therefore 90-one acute angle, gives the other.

THEOREM VI.

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. BE and C-F: and the base of the one BC, will be equal to EF, that of the other. fig. 24.

21

AA

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

LEMMA.

If two sides of a triangle a b c be equal to each other that is, ac= cb; the angles which are opposite to those equal sides, will also be equal to each other; viz. a=b. fig. 11.

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For let the triangle a b c be divided into two triangles a cd, dc 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 d c=dcb; and therefore the angle a=b. Q. E. D.

Cor.

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

THEOREM VII.

fg. 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. fig.

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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, DAĆ+ 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 BAC+CAD or to BAD. Q. E. D.

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

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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 semicircle, and consequently less than a right angle. fig. 26.

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

fig. 27

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Let the lines CA and CB be drawn from the center to the extremities of the chord, then since CA= CB, the angle CABCBA (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 triangle; severally equal to CB, CD, and the angle BCD in the other; therefore (by theo. 6.) A=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.

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