sides of the line AB, make up together two right angles, then CB and BD form one continued right line CD. Corol. 2. Hence, all the angles which can be made, at any point B, by any number of lines, on the same side of the right line CD, are, when taken all together, equal to two right angles. Corol. 3. And, as all the angles that can be made on the other side of the line CD are also equal to two right angles; therefore, all the angles that can be made quite round a point B, by any number of lines, are equal to four right angles. Corol. 4. Hence, also, the whole circumference of a circle, being the sum of the measures of all the angles that can be made about the centre F (def. 57), is the measure of four right angles. Consequently, a semicircle, or 180 degrees, is the measure of two right angles; and a quadrant, or 90 degrees, the measure of one right angle. THEOREM VII. When two lines intersect each other, the opposite angles are equal. For, since the line CE meets the line AB, the two In like manner, the line BE, meeting the line CD, makes the two angles BEC, BED, equal to two right angles. A D Let ABC be a triangle, having the side AB produced to D; then will the outward angle CBD be greater than either of the inward opposite angles A or C. E Therefore, the sum of the two angles AEC, BEC, is equal to the sum of the two BEC, BED (ax. 1). And if the angle BEC, which is common, be taken away from both these, the remaining angle AEC will be equal to the remaining angle BED (ax. 3). And in like manner it may be shown, that the angle AED is equal to the opposite angle BEC. B THEOREM VIII. When one side of a triangle is produced, the outward angle is greater than either of the two inward opposite angles. C F A B D For, conceive the side BC to be bisected in the point E, and draw the line AE, producing it till EF be equal to AE; and join BF. = Then, since the two triangles AEC, BEF, have the side AE the side EF, and the side CE the side BE (by suppos.), and the included or opposite angles at E also equal (th. 7), therefore, those two triangles are equal in all respects (th. 1), and have the angle C the corresponding angle EBF. But the angle CBD is greater than the angle EBF; consequently, the said outward angle CBD is also greater than the angle C. In like manner, if CB be produced to G, and AB be bisected, it may be shown that the outward angle ABG, or its equal CBD, is greater than the other angle A. THEOREM IX. The greater side, of every triangle, is opposite to the greater angle; and the greater angle opposite to the greater side. Let ABC be a triangle, having the side AB greater than the side AC; then will the angle ACB, opposite the greater side AB, be greater than the angle B, opposite the less side AC. For, on the greater side AB, take the part AD equal to the less side AC, and join CD. Then, since BCD is a triangle, the outward angle ADC is greater than the inward opposite angle B (th. 8). But the angle ACD is equal to the said outward angle ADC, because AD is equal to AC (th. 3). Consequently, the angle ACD also is greater than the angle B. And since the angle ACD is only a part of ACB, much more must the whole angle ACB be greater than the angle B. Q. E. D. Again, conversely, if the angle C be greater than the angle B, then will the side AB, opposite the former, be greater than the side AC, opposite the latter. A THEOREM X. For, if AB be not greater than AC, it must be either equal to it, or less than it. But it cannot be equal, for then the angle C would be equal to the angle B (th. 3), which it is not, by the supposition. Neither can it be less, for then the angle C would be less than the angle B, by the former part of this; which is also contrary to the supposition. The side AB, then, being neither equal to AC, nor less than it, must necessarily be greater. Q. E. D. D The sum of any two sides of a triangle is greater than the third side. Let ABC be a triangle; then will the sum of any two of its sides be greater than the third side, as for instance, ACCB greater than AB. For, produce AC till CD be equal to CB, or AD equal to the sum of the two AC+ CB; and join BD:-Then, because CD is equal to CB (by constr.), the angle D is equal to the angle CBD (th. 3). But the angle ABD is greater than the angle CBD, consequently, it must also be greater than the angle D. And, since the greater side of any triangle is opposite to the greater angle (th. 9), the side AD (of the triangle ABD) is greater than the side AB. But AD is equal to AC and CD, or AC and CB, taken together (by constr.); therefore, AC + CB is also greater than AB Q. E. D. THEOREM XI. A D B Corol. The shortest distance between two points, is a single right line drawn from the one point to the other. The difference of any two sides of a triangle, is less than the third side. THEOREMS. Let ABC be a triangle; then will the difference of any two sides, as AB — AC, be less than the third side BC. For, produce the less side AC to D, till AD be equal to the greater side AB, so that CD may be the difference of the two sides AB AC; and join BD. Then, because AD is equal to AB (by constr.), the opposite angles D and ABD are equal (th. 3). But the angle CBD is less than the angle ABD, and consequently also less than the equal angle D. And since the greater side of any triangle is opposite to the greater angle (th. 9), the side CD (of the triangle BCD) is less than the side BC. Q. E. D. Otherwise. Set off upon AB a distance AI equal to AC. Then (th. 20) AC + CB is greater than AB, that is, greater than AI + IB. From these, take away the equal parts, AC, AI, respectively; and there remains CB greater than IB. Consequently, IB is less than CB. Q. E. D. For if they are not equal, one of them must be greater than the other; let it be EFD for instance which is the greater, if possible; and conceive the line FB to be drawn, cutting off the part or angle EFB equal to the angle AEF, and meeting the line AB in the point B. THEOREM XII. When a line intersects two parallel lines, it makes the alternate angles equal to each other. Let the line EF cut the two parallel lines AB, CD; then will the angle AEF be equal to the alternate angle EFD. A A C D F 367 E D Then, since the outward angle AEF, of the triangle BEF, is greater than the inward opposite angle EFB (th. 8); and since these two angles also are equal (by the constr.) it follows, that those angles are both equal and unequal at the same time: which is impossible. Therefore the angle EFD is not unequal to the alternate angle AEF, that is, they are equal to each other. Q. E. D. Corol. Right lines which are perpendicular to one, of two parallel lines, are also perpendicular to the other. THEOREM XIII. When a line, cutting two other lines, makes the alternate angles equal to each other, those two lines are parallel. Let the line EF, cutting the two lines AB, CD, make the alternate angles AEF, DFE, equal to each other; Then will AB be parallel to CD. For if they be not parallel, let some other line, as FG, be parallel to AB. Then, because of these parallels, the angle AEF is equal to the alternate angle EFG (th. 12). But the angle AEF is equal to the angle EFD (by hyp.) Therefore the angle EFD is equal to the angle EFG (ax. 1); that is, a part is equal to the whole, which is impossible. Therefore no line but CD can be parallel to AB. QE D. A E # C F B Corol. Those lines which are perpendicular to the same line, are parallel to each other. THEOREM XIV. When a line cuts two parallel lines, the outward angle is equal to the inward opposite one, on the same side; and the two inward angles, on the same side, are together equal to two right angles. Let the line EF cut the two parallel lines AB, CD; then will the outward angle EGB be equal to the inward opposite angle GHD, on the same side of the line EF; and the two inward angles BGH, GHD, taken together, will be equal to two right angles. For since the two lines AB, CD, are parallel, the angle AGH is equal to the alternate angle GHD, (th. 12). But the angle AGH is equal to the opposite angle EGB (th. 7). Therefore the angle EGB is also equal to the angle GHD (ax. 1). Q. E. D. Again, because the two adjacent angles EGB, BGH, are together equal to two right angles (th. 6); of which the angle EGB has been shown to be equal to the angle GHD; therefore the two angles BGH, GHD, taken together, are also equal to two right angles. A C For, let the line GI be perpendicular to EF. Then will this line be also perpendicular to both the lines AB, CD (corol. th. 12), and consequently the two lines AB, CD, are parallels (corol. th. 13), Q. E. D. G H A C E THEOREM XV. Those lines which are parallel to the same line, are parallel to each other. Let the lines AB, CD, be each of them parallel to the line EF; then shall the lines AB, CD, be parallel to each other. Corol. 1. And, conversely, if one line meeting two other lines, make the angles on the same side of it equal, those two lines are parallels. Corol. 2. If a line, cutting two other lines, make the sum of the two inward angles on the same side, less than two right angles, those two lines will not be parallel, but will meet each other when produced. + E G H B D D F THEOREM XVI. When one side of a triangle is produced, the outward angle is equal to both the inward opposite angles taken together. Let the side AB, of the triangle ABC, be produced to D; then will the outward angle CBD be equal to the sum of the two inward opposite angles A and C. For, conceive BE to be drawn parallel to the side AC of the triangle. Then BC, meeting the two parallels AC, BE, makes the alternate angles C and CBE equal (th. 12). And AD, cutting the same two parallels AC, BF, makes the inward and outward angles on the same side, A and EBD, equal N A B to each other (th. 14). Therefore, by equal additions, the sum of the two angles A and C, is equal to the sun of the two CBE and EBD, that is, to the whole angle CBD (by ax. 2). . E. D. THEOREM XVII. 2 In any triangle, the sum of all the three angles is equal to two right angles. Let ABC be any plane triangle; then the sum of the three angles A+B+C is equal to two right angles. For, let the side AB be produced to D. Then the outward angle CBD is equal to the sum of the two inward opposite angles A+ C (th. 16). To each of these equals add the inward angle B, then will the sum of the three inward angles A+B+C be equal to the sum of the two adjacent angles ABC + CBD (ax. 2). But the sum of these two last adjacent angles is equal to two right angles (th. 6). Therefore also the sum of the three angles of the triangle A+B+C is equal to two right angles (ax. 1). Q. E. D. A Corol. 1. If two angles in one triangle, be equal to two angles in another triangle, the third angles will also be equal (ax. 3), and the two triangles equiangular. B D Corol. 2. If one angle in one triangle, be equal to one angle in another, the sums of the remaining angles will also be equal (ax. 3). Corol. 3. If one angle of a triangle be right, the sum of the other two will also be equal to a right angle, and each of them singly will be acute, or less than a right angle. Corol. 4. The two least angles of every triangle are acute, or each less than a right angle. Let ABCD be a quadrangle; then the sum of the four inward angles, A + B + C + D is equal to four right angles. 3 THEOREM XVIII. In any quadrangle, the sum of all the four inward angles, is equal to four right angles. Let the diagonal AC be drawn, dividing the quadrangle into two triangles, ABC, ADC. Then, because the sum of the three angles of each of these triangles is equal to two right angles (th. 17); it follows, that the sum of all the angles of both triangles, which make up the four angles of the quadrangle, must be equal to four right angles (ax. 2). Q. E. D. Corol. 1. Hence, if three of the angles be right ones, the fourth will also be a right angle. B Corol. 2. And if the sum of two of the four angles be equal to two right angles, the sum of the remaining two will also be equal to two right angles. 4 THEOREM XIX. In any figure whatever, the sum of all the inward angles, taken together, is equal to twice as many right angles, wanting four, as the figure has sides. A A |