CHAPTER II.

Of Straight Lines; first Elements of Rectilineal Figures and equality of Triangles.

§ 1. PROBLEM. To describe an equilateral triangle with a given straight line.

APPLICATION. Let AB, be the given straight line with which it is required to describe an equilateral triangle. (Fig. 10.)

Solution. From the point A, as centre, with the line AB, describe the circle BCD; and from the point B, as centre, with the same line AB, describe the circle ACE; from C, a point of intersection of the circumferences of these two circles draw the straight lines CA, and CB, to the two extremities of the line AB, the triangle ABC, will be the equilateral triangle required.

Proof. The circumferences of the two circles BCD, and ACE, must cut one another, because the centre of the one lies in the circumference of the other, so that the circle must lie partly within and partly without one another; the point C, where these circumferences cut one another being common to both, the straight line AC, is equal to AB, as radii of the circle BCD, and the straight line BC, equal to the AB, as radii of the circle ACE; therefore, the two straight lines AC, and BC, are both equal to AB, therefore they are equal to one another; that is, all the three sides of the triangle ABC, are equal, or, it is equilateral, as was to be proved.

§ 2. PROBLEM. From a given point to draw a straight line equal to a given straight line.

APPLICATION. Let AB, be the given straight line; C, the given point, from which it is required to draw a straight line equal to the given straight line AB. (Fig. 11.)

Solution. Join AC, and with that line describe the equilateral triangle ACD, protract DA, and DC, (indefinitely,) from A, as centre, and with AB, as radius describe the circle BEG, which will cut the protracted DA, because the centre A, lies in it; let this point of intersection be E; from the centre D, with DE, as radius describe the circle EFH, which will cut the protracted DC, because the centre D, lies in it, let this point of intersection be F; the line CF, will be the straight line required, drawn from the point C, and equal to AB.

Proof. The straight lines AE, and AB, are equal as radii of the circle BEG; the sides DC, and DA, are equal as sides of the equilateral triangle ADC; the lines DAE, and DCF, are straight lines being the protractions of the straight lines DA, and DC; the lines DE, and DF, are equal as radii of the same circle EFH; taking from each of the equal straight lines DF, and DE, the equal lines or sides of the triangle DC, and DA, respectively; the remainder CF, will be equal to the remainder AE; the AE, being equal to AB, the CF, is also equal to AB; therefore from the point C, has been drawn a line CF, equal to AB, as was required.

Corol. As from the centre C, with the straight line CF, a circle can be drawn, the line equal to CF, or AB, may be laid in any direction that may be desired.

§ 3. PROBLEM. From the greater of two given straight lines to cut off one equal to the smaller.

APPLICATION. Let AB, be a given straight line larger than CD, it is required to cut off from AB, a line equal to CD. (Fig. 12.)

Solution. From the point A, lay off the straight line AE, in any direction and equal to the given line CD, from the centre A, with AE, as radius describe a circle or section of a circle EFG, which will cut the AB, in F, the line AF, will be the line required cut off from AB, and equal to CD.

Proof. The straight line AE, being made equal to CD; and the AF, being equal to AE, as radius of the same circle, the AF, is equal to ('D, and because CD, is smaller than AB, the circle described with AE, or AF, must cut it, therefore the AE, is cut off from the AB, equal to CD, as was required.

$ 4. THEOREM. If two triangles have two sides in the one equal to two sides in the other, each to each, and the angle between these two sides in the one triangle equal to the angle between the equal sides in the other triangle, the third side of the one will also be equal to the third side of the other; the angles opposite to the equal sides in both triangles will be equal; and the triangles will be equal in all respects.

APPLICATION. Let ABC, and DEF, be two triangles (figure 13,) having the side AB, in the one, equal to the side DE, in the other; the AC, in the first, equal to the DF, in the second, and the angle at A, equal to the angle at D; then will also be: the side BC, equal to the side EF; the angle at B, equal to the angle at E, the angle at C, equal to the angle at F; and the triangles ABC, and DEF, will be equal in all respects.

Demonstration. Lay the triangle ABC, upon the triangle DEF, so that the point A, fall upon the point D, the line AB, upon the line DE; then will the line AC, fall upon DF, because the angle A, is equal to the angle D; the point B, will fall upon the point E, because AB, is equal to DE; the point C, will fall upon the point F, because the line AC, is equal to DF; therefore also the lines CB, and EF, which join these two points will coincide, as they lie between the same points, and two straight lines cannot include a space. Also the angle B, will coincide with the angle E, and the angle C, with the angle F; therefore, the angles opposite to the equal sides will be respectively equal in the two triangles; therefore all the

parts of the two triangles will coincide and the whole triangles will be equal to one another in all respects, as was to be demonstrated.

Corol. This proposition shews that two sides and the included angle determine a triangle, for with these parts a triangle can be made equal to one having these data.

§ 5. THEOREM. In an isosceles triangle the angles opposite to the two equal sides are equal; and if these sides are protracted, the angles which these protractions make with the third side, on the outside of the triangle are also equal.

APPLICATION. Let ABC, (fig. 14,) be an isosceles triangle, in which AB, is equal to AC; then will the angle ABC, be equal to the angle ACB; and when AB, and AC, are protracted on the other side of BC, the angle CBF, will be equal to the angle BCG, outside of the triangle.

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Demonstration. In the protracted AF, take any line AD, greater than AB, and upon the protracted AC, make AE, equal to AD; join BE and CD. In the two triangles ABE, and ACD, the two lines AB, in the one, and AC, in the other, are equal, being the equal sides of the isosceles triangle ABC; the line AE, equal to AD, by construction; and the angle at A, is common to these two triangles; therefore the third side BE, is equal to the third side CD, and the angles of these triangles that are opposite to equal sides, are equal; that is, the angle BEA, equal to the CDA; and EBA, equal to DCA. In the two triangles BCD, and CBE, the sides BD, and CE, are equal, because they are the differences of equal sides, AD, and AB, and of AE, and AC; the angle at D, has been proved equal to the angle at E, therefore these two triangles are equal in all respects, and the angles CBD, and BCE, are equal; these are the angles of the protracted equal sides, with the third side outside of the triangle. Also the angles EBC, and DCB, are equal; therefore, if from each

of the angles ABE, and ACD, which have above been proved to be equal, these equal angles are taken respectively: the remainders ABC, and ACB, will be equal; and these are the angles of the triangle, which are opposite to the equal sides. Therefore, in an isosceles triangle, the angles which the third side makes with the equal sides, are equal, and if the equal sides are protracted, also the angles on the other side of this third side, are equal.

Corol. Every equilateral triangle is also equiangular, because the above demonstration applies to any two sides indiscriminately. § 6. THEOREM. If a triangle has two of its angles equal, the sides opposite to these equal angles will be equal.

APPLICATION. Let ABC, be a triangle in which the angle ABC, be equal to the angle ACB, then will also the side AC, be equal to the side AB, (fig. 15.)

Demonstration. If they are not equal, the one of them must be greater than the other; suppose AB, the greater, let BD, be cut off from BA, and equal to AC; join DC; then in the two triangles ACB, and BDC, there will be the side AC, equal to BD; the side BC, common to both triangles, and the angle ACB, equal to the angle DBC; therefore, the two triangles ABC, and DBC, are equal, and the angles ACB, and DCB, being both equal to the ABC, they are equal to one another; that is, the smaller equal to the greater, which is impossible; therefore, the sides AB, and AC, are not unequal; that is, if two angles in a triangle are equal, the sides opposite to these angles are also equal.

Corol. Every equiangular triangle is also equilateral, because the above reasoning applies to all the three angles equally.

§ 7. THEOREM. On the same side of a straight line there cannot be two different triangles, which have the two sides ending in the same points of that straight line equal to one another.