angle_ABC will co-incide with the whole Triangle DEF, and will be equal thereto; and the remaining † Ax. 8. Angles will co-incide with the remaining Angles t, and will be equal to them, viz. the Angle A B C equal to the Angle D E F, and the Angle A CB equal to the Angle ĎFE. Which was to be demonstrated. PROPOSITION V. THEOREM. equal between themselves : And if the equal Side A B equal to the Side AC; and let the equal Sides AB, AC, be produced directly forwards to D and E. I say the Angle A B C is equal to the Angle ACB, and the Angle CBD equal to the Angle BCE. For affume any Point F in the Line BD, and from * 3 of tbis. A E cut off the Line A G equal * to AF, and join FC, GB. Then because AF is equal to AG, and A B to AC, the two Right Lines FA, AC, are equal to the two Lines GA, A B, each to each, and contain the com+ 4 of this. mon Angle FAG; therefore the Base F C is equal † to the Base G B, and the Triangle AF C equal to equa! 2 equal to the remaining Angles of the other, each to * Ax. 3. Coroll. Hence every Equilateral Triangle is also Equi angular. PROPOSITION VI. . THEOREM. fubtending the equal Angles will be equal be- ET ABC be a Triangle, having the Angle ABC equal to the Angle ACB. I say the Side For if AB be not equal to AC, let one of them, as Therefore if two Angles of a Triangle be equal between themselves, the Sides fubtending the equal Angles are likewise equal between themselves. Which was to be demonstrated. Coroll . Hence every Equiangular Triangle is also Equilateral. PROPOSITION VII. THEOREM. Right Lines equal to two other Right Lines, FOB OR, if it be possible, let two Right Lines AD, DB, equal to two others A C, CB, each to each, be constituted at different Points C and D, towards the same Parts CD, and having the same Ends A and B which the first Right Lines have, so that CA be equal to AD, having the fame End A which CA hath; and CB equal to D B, having the same End B; and let CD be joined. Then because AC is equal to AD, the Angle * 5 of this. A CD will be equal * to the Angle ADC, and con sequently the Angle ADC is greater than the Angle BCD; wherefore the Angle BDC will be much greater than the Angle B CD. Again, because CB is equal to D B, the Angle BDC will be equal to the Angle BCD, but it has been proved to be much greater, which is impossible. Therefore on the same Right Line cannot be constituted two Right Lines equal to two ather Right Lines, each to each, at different Points, on the same Side, and having the fame Ends which the first Right Lines have; which was to be demonstrated. PRO PROPOSITION VIII. THEOREM. to two Sides of the other, each to each, and DF, each to each, viz. A B equal to DE, and AC For if the Triangle ABC be applied to the Trian- of thisa PRO PROPOSITION. IX. PROBLEM. To cut a given Right-lin'd Angle into two equal Parts. L ET BAC be a given Right-lin'd Angle, which is required to be cut into two equal Parts. Affume any Point D in the Right Line A B, and 3 of tbis. cut off A E from the Line AC equal to AD*; join + 1 of tbis. D E, and thereon make † the Equilateral Triangle DEF, and join AF. I say, the Angle BAC is cut into two equal Parts by the Line AF. For because A D is equal to A E, and A F is common, the two Sides D A, AF, are each equal to the two Sides A E, AF, and the Base D F is equal to 1 8 of this, the Base E F; therefore I the Angle DAF is equal to the Angle E AF. Wherefore a given Right-lin'd Angle is cut into two equalParts; which was to be done. PROPOSITION X. PROBLEM. To cut a given finite Right Line into two equal Parts. L be cut into two equal Parts. I of this. Upon it make * an Equilateral Triangle A B C, and te of tbis. bisect † the Angle ACB by the Right Line CD. I say, the Right Line A B is bifected in the Point D. For because AC is equal to CB, and CD is common, the Right Lines AC, CD, are each equal to the two Right Lines BC, CD, and the Angle ACD | 4 of tbii. equal to the Angle BCD; therefore I the Base AD, is equal to the Bale DB. And fo the Right Line AB is bisected in the Point D; which was to be done, PRO |