PROPOSITION VIII. THEOREM. If two Triangles have two Sides of the one equal to two Sides of the other, each to each, and the Bafes equal, then the Angles contained under the equal Sides will be equal. LET ET the two Triangles be ABC, DEF, having two Sides A B, AC, equal to two Sides DE DF, each to each, viz. AB equal to D E, and AC to DF; and let the Bafe B C be equal to the Base EF. I fay, the Angle BAC is equal to the Angle EDF. For if the Triangle ABC be applied to the Triangle DEF, fo that the Point B may co-incide with E, and the Right Line BC with EF, then the Point C will co-incide with F, because BC is equal to EF. And fo fince B C co-incides with E F, BA and AC will likewife co-incide with E D and D F. For if the Bafe B C should co-incide with E F, and at the fame time the Sides BA, AC, fhould not co-incide with the Sides ED, DF, but change their Position, as EG, GF, then there would be constituted on the fame Right Line two Right Lines equal to two other Right Lines, each to each, at feveral Points, on the fame Side, having the fame Ends. But this is proved to be otherwife; therefore it is impoffible for the 7 of this. Sides BA, AC, not to co-incide with the Sides ED, DF, if the Bafe B C co-incides with the Base EF; wherefore they will co-incide, and confequently the Angle B A C will co-incide with the Angle EDF, and will be equal to it. Therefore if two Triangles have two Sides of the one equal to two Sides of the other, each to each, and the Bafes equal, then the Angles contained under the equal Sides will be equal; which was to be demonstrated. PRO +3 of this 1 of this. PROPOSITION IX. PROBLEM. To cut a given Right-lin❜d Angle into two equal Parts. 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 cut off AE from the Line AC equal to AD+; join DE, and thereon make the Equilateral Triangle DEF, and join AF. I fay, the Angle BAC is cut into two equal Parts by the Line AF. For because AD is equal to A E, and AF is common, the two Sides DA, AF, are each equal to the two Sides A E, AF, and the Bafe DF is equal to I of this. the Bafe EF; therefore the Angle DAF is equal to the Angle EAF. Wherefore a given Right-lin❜d Angle is cut into two equal Parts; which was to be done. 1 of this. PROPOSITION. X. PROBLEM. To cut a given finite Right Line into two equal Parts, LETAB be a given finite Right Line, required to be cut into two equal Parts. Upon it make an Equilateral Triangle ABC, and #9 of this. bifect the Angle AC Bby the Right Line CD. Í fay, the right Line AB is bifected in the Point D. For because AC is equal to CB, and CD is common, the Right Lines A C, CD, are each equal to the two Right Lines BC, CD, and the Angle ACD 4 of this, equal to the Angle BCD; therefore the Bafe AD, is equal to the Bafe DB. And fo the Right Line AB is bifected in the Point D; which was to be done, I * PRO, PROPOSITION XI. PROBLEM. To draw a Right Line at Right Angles to a given Right LETAB be the given Right Line, and C the given the Point C, at Right Angles to A B. Affume any Point D in AC, and make C E equal *to CD, and upon DE make † the Equilateral Tri- * 3 of this. angle FDE, and join FC. I fay, the Right Linet FC is drawn from the Point C, given in the Right Line AB at Right Angles to A. B. of this. For because DC is equal to CE, and FC is common, the two Lines DC, CF, are each equal to the two Lines E C, CF; and the Bafe DF is equal to the Bafe F E. Therefore the Angle D C F, is equal to the Angle ECF; and they are adjacent Angles. But when a Right Line, ftanding upon a Right Line, makes the adjacent Angles equal, each of the equal Angles is a Right Angle; and confequently DCF, Def. 10, FCE, are both Right Angles. Therefore the Right Line FC, &c. which was to be done. PROPOSITION XII. PROBLEM. To draw a Right Line perpendicular, upon a given in- LET AB be the given infinite Line, and C the Affume any Point D on the other fide of the Right * Per 10 of this. Perpendicular CH on the given infinite Right Line For because GH is equal to HE, and HC is common, GH and HC are each equal to EH and HC, and the Base CG is equal to the Bafe CE. Therefore Def. 10. the Angle CHG is equal to the Angle CHE; and they are adjacent Angles. But when a Right Line, standing upon another Right Line, makes the Angles equal between themselves, each of the equal Angles Def. 10. is a Right one*, and the said standing Right Line is call'd a Perpendicular to that which it ftands on. Therefore GH is drawn perpendicular, upon a given infinite Right Line, from a given Point out of it; which was to be demonstrated. PROPOSITION XIII. THEOREM. When a Right Line, standing upon a Right Line, makes Angles, thefe fhall be either two Right Angles, or together equal to two Right Angles. FOR let a Right Line AB, ftanding upon the Right Line CD, make the Angles CBA, ABD. Ifay, the Angles CBA, ABD, are either two Right Angles, or both together equal to two Right Angles. * Def. 10. For if CBA be equal to A BD, they are each of +11 of this. them Right Angles: But if not, draw BE from the Point B, at Right Angles to CD. Therefore the Angles CBE, EBD, are two Right Angles: And because CBE is equal to both theAngles CBA, ABE, add the Angle EBD, which is common; and the two 4x. 2. Angles CBE, EBD, together, are equal to the three Angles CBA, ABE, EBD, together. Again because the Angle DBA is equal to the two Angles DBE, EBA, together, add the common Angle ABC, and the two Angles DBA, ABC, are equal to the three Angles DBE, EBA, ABC, together. But it has been prov'd that the two Angles CBE, EBD, together, are likewife equal to there three Angles: But Things that are equal to one and the fame, * Ax. 1. are equal between themselves. Therefore likewife the Angles CBE, EBD, together, are equal to the * 2 Angles Angles DBA, ABC, together; but CBE, EBD, are two Right Angles. Therefore the Angles DBA, ABC, are both together equal to two Right Angles. Wherefore when a Right Line, ftanding upon another Right Line, makes Angles, these shall be either two Right Angles, or together equal to two Right Angles; which was to be demonstrated. PROPOSITION XIV. THEOREM. If to any Right Line, and Point therein, two Right Lines be drawn from contrary Parts, making the adjacent Angles, both together, equal to two Right Angles, the faid two Right Lines will make but one Straight Line. FOR let two Right Lines BC, BD, drawn from contrary Parts to the Point B, in any Right Line AB, make the adjacent Angles ABC, ABD, both together, equal to two Right Angles. I fay, BC, BD, make but one Right Line. For if BD, CB, do not make one straight Line, let CB and B'E make one. * Then, because the Right Line AB ftands upon the Right Line CBE, the Angles ABC, ABE, together, will be equal to two Right Angles. But the Angles 13 of this. ABC, ABD, together, are alfo equal to two Right Angles. Now taking away the common Angle ABC, and the remaining Angle ABE is equal to the remaining Angle ABD, the lefs to the greater, which is impoffible. Therefore BE, BC, are not one straight Line. And in the fame Manner it is demonftrated, that no other Line but BD is in a ftraight Line with CB; wherefore CB, BD, fhall be in one ftraight Line. Therefore if to any Right Line, and Point therein two Right Lines be drawn from contrary Parts, making the adjacent Angles, both together, equal to two Right Angles, the faid two Right Lines will make but one ftraight Line; which was to be demonstrated. PRO |