to Proposition fourth, the angles on the other side of the base are equal. It is evident that some line will bisect the vertical angle; and although the method of doing it is not known till Problem 9 be solved, yet this does not affect the truth of the demonstration. PROPOSITION VI. THEOREM. If two angles of a triangle be equal to one another, the sides which subtend, or are opposite to, those angles, shall also be equal to one another. Given, ABC a triangle having the angle ABC equal to the angle ACB, to prove that the side AB is also equal to the side AČ. Gal(Const.) For, if AB be not equal to AC, one of them is greater than the other. Let AB be the greater, and from it cut (1. 3) off D B equal to AC, the less, and join DC. (Dem.) Because in the triangles DBC and ACB, DB is equal to AC, and BC common to both, the two sides DB and BC are equal to the two AC and CB, each to each; but the angle DBC is also equal to the angle ACB; awtherefore the base DC is equal to the base AB, and the triangle DBC is equal to the triangle ACB (1. 4), the less equal to the greater; which is absurd. Therefore AB is not greater than AC, and in the same manner it may be shewn that AB is not less than AC; that is, AB is equal to AC. COR.-Hence every equiangular triangle is also equilateral. PROPOSITION VII. THEOREM. Upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity, equal to one another... Given, two triangles ACB and ADB, upon the same base AB, and upon the same side of it, which have their sides CA and DA, terminated in A equal to one another, to prove that all their sides CB and DB, terminated in B, cannot be equal to one another.. · Join CD; then, in the case in which the vertex of each of the triangles is without the other triangle, because AC is equal to AD, the angle ACD is equal (I. 5) to the angle ADC. But the angle ACD is greater than the angle BCD; therefore the angle ADC is greater also than BCD; still greater, then, is the angle BDC than the angle BCD. Again, if CB were equal to DB, the angle BCD would be equal to the angle BDC (I. 5), but the angle BDC has been shewn to be greater than the angle BCD; therefore BC is not equal to BD. But if one of the vertices, as D, be within the other triangle ACB; produce AC, AD, to E, F; then, because AC is equal to AD in the triangle ACD, the angles ECD and FDC, upon the other side of the base CD, are equal to one another, but the angle ECD is greater than the angle BCD; wherefore the angle FDC is likewise greater than BCD; much more then is the angle BDC greater than the angle BCD. Again, if CB were equal to DB, the angle BDC would be equal to the a angle BCD; but the angle BDC has been proved to be greater than the angle BCD; therefore the side BC is not equal to BD. The case in which the vertex of one triangle is upon a side of the other, needs no demonstration. PROPOSITION VIII. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal, the angle which is contained by the two sides of the one shall be equal to the angle contained by the two sides of the other. Or, if the three sides of one triangle be respectively equal to those of another, the triangles are equal in every respect, and have the angles equal that are opposite the equal sides. Given, ABC and DEF, two triangles having the two sides AB and AC, equal to the two sides DE and DF, each to eachnamely, AB to DE, and AC to DF; and also the base BC equal to the base EF; to prove that the angle BAC is equal to the angle EDF. (Const.) For, if the triangle ABC be applied to the triangle DEF, so that the point B be on E, and the straight line BC upon EF, (Dem.) the point C shall also coincide with the point F, because BC is equal to EF; and BC coinciding with EF, there. fore BA and AC shall coincide with ED and DF; for, if BA and CA do not coincide with ED and FD, but have a different situation as EG and FG, then, upon the same base EF, and upon the same side of it, there can be two triangles EDF and EGF, that have their sides which are terminated in one CE extremity of the base equal to one another, and likewise their sides terminated in the other extremity. But this is impossible (I. 7); therefore if the base BC coincides with the base EF, the sides BA and AC cannot but coincide with the sides ED and DF; wherefore, likewise, the angle BAC coincides with the angle EDF, and is equal (Ax. 8) to it. *** See Appendix-Proposition (A.) PROPOSITION IX. PROBLEM. To bisect a given rectilineal angle—that is, to divide it into two equal angles. Given, the rectilineal angle BAC; it is required to bisect it. (Const.) Take any point D in AB, and from AC cut (I. 3) off AE equal to AD; join DE, and upon it describe (I. 1) an equilateral triangle DEF; then join AF; the straight line AF bisects the angle BAC. (Dem.) Because AD is equal to AE, and AF is common to the two triangles DAF and EAF; the two sides DA and AF are equal to the two sides EA and AF, each to each ; .but the base DF is also equal to the base B/ c EF; therefore the angle DAF is equal (1. 8) to the angle EAF; wherefore the given rectilineal angle BAC is bisected by the straight line AF. PROPOSITION X. PROBLEM. To bisect a given finite straight line-that is, to divide it into two equal parts. Given, the straight line AB; it is required to divide it into two equal parts. , (Const.) Describe (I. 1) upon it an equilateral triangle ABC, and bisect (I. 9) the angle ACB by the straight line CD. AB is cut into two equal parts in the point D. (Dem.) Because AC is equal to CB, and CD common to the two triangles ACD and BCD; the two sides AC and CD are equal to the two BC and CD, each to each; but the angle ACD is also equal to the angle BCD; therefore the base AD is equal to the base DB (I. 4), and the * straight line AB is divided into two equal parts in the point D. PROPOSITION XI. PROBLEM. To draw a straight line at right angles to a given straight line, from a given point in the same. Given, a straight line AB, and a point C in it; it is required to draw a straight line from the point. C at right angles to AB. , (Const.) Take any point D in AC, and (I. 3) make CE equal to CD, and upon DE describe (I. 1) the equilateral triangle DFE, and join FC; the straight line FC, drawn from the given point C, is at right angles to the given straight line AB. (Dem.) Because DC is equal to CE, and FC common to the two triangles DCF, AD Ć E B ECF; the two sides DC and CF are equal to the two EC and CF, each to each; but the base DF is also equal to the base EF; therefore the angle DCF is equal (I. 8) to the angle ECF; and they are adjacent angles; therefore (Def. 11) each of the angles DCF and ECE is a right angle. Wherefore, from the given point C, in the given straight line AB, FC has been drawn at right angles to AB. PROPOSITION XII. PROBLEM. Portory To draw a straight line perpendicular to a given straight line of an unlimited length, from a given point without it. Given the straight line AB, which may be produced to any length both ways, and a point C without it. It is required to draw a straight line perpendicular to AB from the point C. (Const.) Take any point D upon the other side of AB, and from the centre C, with the radius CD, describe (Post. 3) the circle EGF meeting AB in F and G; bisect (I. 10) FG in A F G B H, and join CF, CH, and CG; the straight line CH, drawn from the given point C, is perpendicular to the given straight line AB. (Dem.) Because FH is equal to HG, and HC common to the two triangles FHC and GHC, the two sides, FH and HC,, are equal to the two, GH and HC, each to each; now, the base CF is also equal (Def. 17) to the base CG; therefore the. angle CHF is equal (I. 8) to the angle CHG; and they are adjacent angles; therefore each of them is a right angle,. ? and CH is perpendicular to AB (Def. 11), and it is drawn from 5 the point C, which was required to be done. PROPOSITION XIII. THEOREM. The angles which one straight line makes with another upon ? one side of it, are either two right angles, or are together equals to two right angles. 5. Given the straight line AB making with CD, upon one side of it, the angles CBA and ABD; to prove that these are either two right angles, or are together equal to two right angles. E A (Dem.) For, if the angle CBA be equal to ABD, each of them is a right angle (Def. 11); but, if not, (Const.) from the point B draw BÈ at right angles (I. 11) to CD; therefore the D B C D B C angles CBE and EBD are two right angles; and because CBE is equal to the sum of the two angles CBA and ABE, add the angle EBD to each of these equals; therefore the two angles CBE and EBD are equal (AX. 2) to the three angles CBA, ABE, and EBD. Again, because the angle DBA is equal to the two angles DBE and EBA, add to these equals the angle ABC; therefore the two angles DBA and ABC are equal to the three angles DBE, EBA, and ABC; but the two angles CBE and EBD have been demonstrated to be equal to the same three angles; and things that are equal to the same are equal (Ax. 1) to one another; therefore the angles CBE and EBD are equal to the angles DBA and ABC; but CBE and EBD are two right angles; therefore DBA and ABC are together equal to two right angles. The truth of this proposition may be easily shewn as follows: The angle DBA is greater than a right angle by the angle EBA, and the angle ABC is less than a right angle by the same angle EBA; therefore their sum is two right angles. COR. Hence if one of a pair of supplementary angles be equal to one of another pair, the remaining angles must be equal. PROPOSITION XIV. THEOREM. If, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line. Given that at the point B in the straight line AB, the two straight lines BC and BD, upon the opposite sides of AB, make the adjacent angles ABC . A and ABD equal together to two right angles. To prove that BD is in the same straight line with CB. (Const.) For, if BD be not in the same straight line with CB, let BE be in the C BS same straight line with it; then (Dem.) because the straight line AB makes angles with the straight line |