Any triangle can have but one right angle. COROLLARY 1. COROLLARY 2. In any triangle, if one angle be right, the other two are together equal to a right angle; and if one angle be equal to the other two, it is a right angle. COROLLARY 3. If two triangles have two angles in the one respectively equal to two angles in the other, the remaining angles are also equal. COROLLARY 4. In a right-angled isosceles triangle, each angle at the base is half a right angle. COROLLARY 5. Each angle of an equilateral triangle is equal to a third part of two right angles, or to two-thirds of one right angle. COROLLARY 6. From the foregoing corollary may be derived a method of trisecting a right angle. Upon any portion of the side CB construct an equilateral triangle CDB (a), and bisect the angle CBD by the line EB (b); then is the right angle ABC divided into three equal parts; for the whole angle CBD being equal to two-thirds of ABC (c), its halves are each equal to one-third, and the angle ABD is the remaining third part. By successive bisections of the angles ABD, DBE, and EBČ, the right angle ABC may be divided into 6, 12, 24, &c., equal parts. COROLLARY 7. All the internal angles of any rectilineal figure (ABCDE), together with four right angles, are equal to twice as many right angles as the figure has sides. DEMONSTRATION. Take any point F within the figure, and draw the straight lines FA, FB, FC, FD, FE. There are formed as many triangles as the figure has sides, therefore all their angles taken together are equal to twice as many right angles as the figure has sides (a); but the angles at the point F are together equal to four right angles (b), therefore all the internal angles of the figure ABCDE, together with four right angles, are equal to twice as many angles as the figure has sides. SCHOLIA. 1. It should be observed, that if the figure has a re-entrant angle, as ABC, the internal angle (although greater than two right angles) must be taken, and not the external angle. COROLLARY 8. The external angles of any rectilineal figure (ABC) are together equal to four right angles. DEMONSTRATION. For each external angle ABD, with the adjacent internal angle ABC, is equal to two right angles (a); therefore all the external angles, together with all the internal angles, are equal to twice as many right angles as the figure has sides; but the internal angles, together with four right angles, are equal to twice as many right angles as the figure has sides (b); take away from both the internal angles, and the external angles remain, equal to four right angles (c). B E A (a) I. 1. (a) I. 32 B. (a) I. 13. SCHOLIA. 2. If the figure has a re-entrant angle, as ABC, the angle DBC, or the difference between the angle ABC and two right angles, must be deducted from the sum of all the exterior angles. Thus the sum of the angles LAB, MCE, FEG, HGI, KIA, less the angle DBC, will equal four right angles. 3. The following theorem is introduced as depending upon the preceding, and because of its use in explaining the principles of construction of the quadrant and sextant. A... M THEOREM. If an exterior angle (BCD) of a triangle (ABC) be bisected, and also one of the interior and opposite angles (BAC), the angle (E) formed by the bisecting lines is equal to half the other interior and opposite angle (B) of the triangle. B (a) I. 16. (b) Theor. attached DEMONSTRATION. The lines AE and CE will meet; for because the angle BAC is less than the exterior angle BCD (a), the half of the former EAC is less than the half of the latter ECD; add ECA to both, and EAC and ECA are less than ECD and ECA; but ECD and ECA are together equal to two right angles, therefore EAC and ECA are together less than two right angles, and therefore the lines AE and CE must meet on the same side of AC as those angles (b); let them meet in E. Then, because ECD is the exterior angle of the triangle EAC, it equals the two interior angles EAC and E (c), and therefore twice ECD equals twice the angles EAC and E (d), that is, the angle BCD is equal to the sum of the angle BAC and twice the angle E; but the angle BCD is equal to the two interior angles BAC and B (c), therefore the sum of BAC and twice E is equal to the sum of BAC and B (e), and taking BAC from both, therefore the angle B is equal to twice the angle É (ƒ). PROPOSITION XXXIII. to I. 29. (c) I. 32 A. (d) Ax. 6. (e) Ax. 1. (f) Ax. 3. THEOREM.-If straight lines (AC and BD) join the adjacent extremities of two equal and parallel straight lines (AB and CD), they are themselves equal and parallel. CONSTRUCTION. Draw the diagonal BC. DEMONSTRATION. Because in the triangles ABC and DBC the sides AB and CD are equal (a), the side BC common to both, and the angle ABC equal to the alternate angle BCD (), therefore the two sides AC and BD are equal (c), and also the two angles ACB and CBD are equal (c); and because the straight line BC intersects the straight lines AC and BD, and forms alternate angles ACB and CBD equal to each other, therefore AC is parallel to BD (d). B (a) Hypoth. PROPOSITION XXXIV. THEOREM.-If a figure (ACDB) be a parallelogram, [1] the opposite sides are equal to one another (AB to CD and AC to BD), [2] as are also the opposite angles (A to D and ACD to DBA), [3] and the parallelogram is bisected by its diagonal (CB). DEMONSTRATION. Because AB is parallel to CD (a), the angle ABC is equal to the alternate angle BCD (6); and because AC is parallel to BD (a), the angle ACB is equal to the alternate angle DBC (6). Then, because in the two triangles ABC and DBC the two angles ABC and ACB are equal to BCD and DBC, and the side BC common to both, [1] therefore the sides AB and AC are equal to the opposite sides CD and BD (c), the angle A is equal to the angle D (c), and the triangle ABC is equal to the triangle DBC (d), [3] therefore the parallelogram is bisected by the diagonal BC. (a) Def. 26. [2.] Also because the angle ABC is equal to BCD, and the angle ACB is equal to DBC, therefore the whole angle ACD is equal to the opposite whole angle ABD (e), and the angle Ä has been proved to be equal to the opposite angle D. SCHOLIUM. This theorem consists of three distinct propositions, which are here separately distinguished by numbers. The converse propositions to the first and second may be demonstrated as follows. THEOREM. If in any four-sided figure (ABDC) the opposite sides are equal, it is a parallelogram. DEMONSTRATION. Draw BC. Then, because in the triangles ABC and DBC the sides AC and AB are respectively equal to the sides BD and CD (a), and BC is common to both, therefore the angle ABC is equal to the angle BCD, and the angle ACB to the angle CBD (b). And because the alternate angles ABC and BCD are equal, therefore the straight lines AB and CD are parallel; also the alternate angles ACB and CBD being equal, the straight lines AC and BD are parallel (c), and therefore the figure ABDC is a parallelogram (d). THEOREM. If in any four-sided figure (ABDC) the opposite angles are equal, it is a parallelogram. DEMONSTRATION. Because the angle A is equal to D, and the angle B to C (a), therefore A and B are equal to D and C (b), and the four angles are together double of A and B; but the four angles together are equal to four right angles (c), therefore A and B are equal to two right angles, and the line AC is parallel to BD (d); and in a similar manner it may be proved that AB is parallel to CD, therefore the figure ABDC is a parallelogram (e). B (a) Hypoth. (b) I. 8. (c) I. 27. (d) Def. 26. A D C (a) Hypoth. COROLLARY 1. If a parallelogram (ABDC) have one right angle, all its angles are right angles. For since the adjacent angles A and B are together equal to two right angles (a), if one of them is a right angle (b), the other must also be a right angle; and the angles C and D are right angles, being equal to their opposite angles B and A (c). (a) I. 29. COROLLARY 2. If two parallelograms have an angle of the one equal to an angle of the other, the remaining angles shall be respectively equal. For the angles opposite the equal angles are equal to them (a), and therefore to each other (b); and the angles adjacent to the equal angles are equal, being, with these equals, equal to two right angles (c). COROLLARY 3. The diagonals of a parallelogram bisect each other. For since, in the triangles AEC and DEB, the side AC is equal to the side BD (a), the angle EAC to EDB, and the angle ECA to EBD (b), therefore the side AE is equal to ED, and the side CE to EB (c). SCHOLIUM. By means of the thirty-fourth proposition we may divide a finite straight line into any given number of equal parts in the following manner. Let AB be the given line. Draw AC, making any angle with AB; take any part AD, and make DE, EC equal to it until as many equal parts have been taken as that into which the line AB is to be divided. Join CB, and draw EG, DF parallel to CB, then the line AB shall be divided into the required number of equal parts. Draw DH and EI parallel to AB; then they are also parallel to each other (a); therefore the angles CEI, EDH, and DAF are equal to each other (b), and also the angles C, DEH, and ADF are equal to each other (b), and the sides CE, ED, and DA being equal (c), the sides EI, DH, and AF are also equal (d). Then, because FG and GB are equal to DH and EI (e), therefore they are equal to AF (ƒ). PROPOSITION XXXV. (a) I. 34. (b) Ax. 1. (c) I. 29. (a) Hypoth. I. 30. (b) I. 29. Ax. 1. THEOREM.—If parallelograms (ABCD and EBCF) are upon the same base and between the same paral lels, they are equal in area. CONSTRUCTION. Produce the side BC to G. DEMONSTRATION. Because the lines AB and DC are parallel (a), the angle DCG is equal to the internal angle ABG (6); and Def. 26. I. 29. because EB and FC are parallel (a), the angle FCG is equal to the internal angle EBG (6); therefore the angle DCF is equal to the angle ABE (c). Then, because in the triangles DCF and ABE the two sides DC and CF are equal to the two sides AB and BE (d), and the angle DCF is equal to ABE, therefore the triangles themselves are equal (e). Then, from the quadrilateral ABCF take away the triangle DCF, the remainder is the parallelogram ABCD; and from the same quadrilateral take away the triangle ABE, the remainder is the parallelogram EBCF; therefore the parallelograms ABCD and EBCF are equal in area (c). Def. 26. (d) I. 34. (e) I. 4. SCHOLIA. 1. The demonstration of Euclid has not been adhered to in this proposition, in order to avoid its division into distinct cases, according to the relative positions of the lines AD and EF. 2. The word "equal" is, in this and several following propositions, used by Euclid in a sense somewhat different to that in which it has been employed in the previous portion of the Elements; namely, to denote simply equality in area, and not absolute sameness in form. In order to avoid any ambiguity, the phrase "equal in area" has been substituted for equal; and the latter is only used to signify perfect identity of form. PROPOSITION XXXVI. THEOREM.-If parallelograms (ABCD and EFGH) are upon equal bases and between the same parallels, they are equal to one another in area. CONSTRUCTION, Draw BE and CH. DEMONSTRATION. Because the lines BC and EH are equal to the same FG (a), they are equal to one another (b); but they are also parallel (c); therefore BE and CH, which join their extremities, are parallel (d), and EBCH is a parallelogram equal in area both to ABCD and EFGH (e); and therefore the parallelograms ABCD and EFGH are equal in area (b). (a) Hypoth. and |