AXIOMS. 1. Things which are equal to the same thing, are equal to one another. 2. If equals be added to equals, the wholes are equal. 3. If equals be taken from equals, the remainders are equal. 4. If equals be added to unequals, the wholes are equals. 5. If equals be taken from unequals, the remainders are unequals. 6. Things which are double of the same, are equal to one another. 7. Things which are halves of the same, are equal to one another. 8. Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another. 9. The whole is greater than its part. 10. All right angles are equal to one another. POSTULATES. 1. That a right line may be drawn from any one given point to another. 2. That a right line may be produced or continued at pleasure. 3. That from any centre and with any radius the circumference of a circle may be described. 4. It is also required, that the equality of lines and angles to others given, be granted as possible: That it is possible for one right line to be perpendicular to another, at a given point or distance; and that every magnitude has its half, third, fourth, &c., part. SIGNS. The Sign = denotes the quantities between which it stands to be equal. The Sign + denotes the quantity it precedes to be added. The Sign - denotes the quantity which it precedes to be subtracted. To denote that four quantities, A, B, C, D, are proportional, they are usually written thus: A: B:: C: D, and read thus: A is to B, so is C to D; but when three quantities, A, B, C, are proportional, the middle quantity is repeated, and they are written A: B:: B: C. CHAPTER II. PRACTICAL GEOMETRY. GEOMETRICAL THEOREMS. THEOREM I. If a right line falls on another, as AB, or EB does on CD, it either makes with it two right angles, or two angles equal to two right angles. 1. If AB be perpendicular to CD, then the angles CBA and ABD, will be each a right angle. 2. But if EB fall obliquely on CD, then will the angles DBE + EBC = DBE + EBA (= DBA) + ABC, or to two right angles. Q. E. D. Corollary 1. If any number of right lines are drawn from one point on the same side of a right line; all the angles made by these lines will be equal to two right angles. Cor. 2. All the angles which can be made about a point, will be equal to four right angles. THEO. II. If one right line crosses another, as AC does BD, the opposite angles, made by those lines, will be equal to each other, that is AEB to CED and BEC to AED. By Theorem I. BEC + CED = 2 right angles, and CED + DEA = 2 right angles. Therefore (by Axiom 1.) BEC + B CED = CED + DEA, take CED from both, and there remains BEC = DEA. (by Axiom 3.) E In the same manner CED + AED = 2 right angles; and AED + AEB = 2 right angles: wherefore taking AED from both, there remains CED= AEB. Q. E. D. = THEO. III. If a right line crosses two parallel lines as GH does AB and CD then, 1st. The sum of the interior angles on the same side will be equal to two right angles, that is AEF + CFE equal to two right angles and BEF + DFE equal to two right angles. 2nd. The alternate angles will be equal, that is AEF = EFD and BEF — CFE. 3rd. The exterior angle will be equal to the interior and opposite one on the same side, that is AEG CFE and BEG =DFE. B 1st. If the angles AEF and CFE be not equal to two right angles, let them, if possible, be greater than two right angles;-then because the lines AE and CF are not more parallel than EB and FD, the angles BEF and DFE are also greater than two right angles. Therefore the four angles AEF, CFE, BEF, DFE are greater than four right angles and (by Theorem I.) they are also equal to four right angles which is absurd. /H In the same manner it may be shown that the angles AEF and CFE cannot be less than two right angles. Therefore Again (by Therefore they are equal to two right angles. Wherefore also the angles BEF and DFE are equal to two right angles. 2nd. Now AEF + CFE = two right angles. Theorem I.) CFE + DFE = two right angles. (by Axiom 1) AEF + CFE = CFE + DFE. Take away CFE from both and there remains AEF DFE (by Axiom 3). In the same manner we prove that BEF = CFE. 3rd. Now BEF = CFE, and (by Theorem II.) BEF = AEG. Therefore AEG = CFE (by Axiom 1.) In the same way we prove BEG DFE. Q. E. D. THEO. IV. If in any triangle ABC, one of its sides as BC is produced towards D, it will make the external angle ACD equal to the two internal opposite angles taken together; viz. to B and A. Through C let CE be drawn parallel to AB; then since BD cuts the two parallel lines, BA, CE; the angle ECDB, (by part 3 of the last Theo.) and again, since AC cuts the same parallels, the angle ACEA (by part A 2 of the last Theo.) Therefore ECD + ACE = ACD B+ A. Q. E. D. = THEO. V. In any triangle ABC, all the three angles taken together are equal to two right angles, viz. A + B + ACB = two right angles. Produce BC to any distance as D, then (by the last Theo.) ACD = B+ A; to both add ACB; then ACD + ACB = A + B + ACB: But ACD + ACB 2 right angles (by Theo. = I.), therefore the three angles A + B = Cor. 1. If one angle of a triangle be known, the sum of the other two is also known: For since the three angles |