gles of the figure are equal to twice as many right angles as the figure has sides, wanting four. COR. 1. All the exterior angles of any rectilineal figure are together equal to four right angles. Because every interior angle ABC, with its adjacent exterior ABD, is equal (6. 1.) to two right angles; therefore all the interior, together with all the exterior angles of the figure, are equal to twice as many right angles as there are sides of the figure; that is, they are equal to all the interior angles of the figure, together with four right angles; therefore all the exte rior angles are equal to four right angles. COR. 2. The sum of the angles in a quadrilateral is equal to two right angles multiplied by 4 — 2, which amounts to four right angles ; hence, if all the angles of a quadrilateral are equal, each of them will be a right angle; a conclusion which sanctions the Definitions 24 and 25, where the four angles of a quadrilateral are said to be right, in the case of the rectangle and the square. COR. 3. The sum of the angles of a pentagon is equal to two right angles multiplied by 5-2, which amounts to six right angles; hence, when a pentagon is equiangular, each angle is equal to the fifth part of six right angles, or g of one right angle. COR. 4. The sum of the angles of a hexagon is equal to 2 × (6—2), or eight right angles; hence, in the equiangular hexagon, each angle is the sixth part of eight right angles, or of one right angle. SCHOLIUM. When this proposition is applied to polygons, which have re-entrant angles, as ABC, each re-entrant an gle must be regarded as greater than two right angles. And, by joining BD, BE, BF, the figure is divided into four triangles, which contain eight right angles; that is, as many times two right angles as there are units in the number of sides diminished by two. But to avoid all ambiguity, we shall henceforth limit our reasoning to polygons with salient angles, which might otherwise be named convex polygons. Every convex po A E F lygon is such that a straight line, drawn at pleasure, cannot meet the contour of the polygon in more than two points. The straight lines which join the extremities of two equal and parallel straight lines, towards the same parts, are also themselves equal and parallel. Let AB, CD, be equal and parallel straight lines, and joined towards the same parts by the straight lines AC, BD; AC, BD are also equal and parallel. Join BC; and because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal (21. 1.); and because AB is equal to CD, and BC common to the two triangles ABC, DCB, the two sides AB, BC are equal to the two DC, CB; and the angle ABC is equal to C B D the angle BCD; therefore the base AC is equal (1. 1.) to the base BD, and the triangle ABC to the triangle BCD, and the other angles to the other angles (1. 1.) each to each, to which the equal sides are opposite; therefore the angle ACB is equal to the angle CBD; and because the straight line BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD equal to one another, AC is parallel (19. 1.) to BD; and it was shewn to be equal to it. COR. 1. Hence, if two opposite sides of a quadrilateral are equal and parallel, the remaining sides will also be equal and parallel, and the figure will be a parallelogram. COR. 2. And every quadrilateral, whose opposite sides are equal, is a parallelogram, or has its opposite sides parallel. For, having drawn the diagonal BC; then, the triangles ABC, CBD, being mutually equilateral (hyp.), they are also mutually equiangular (Th. 5.), or have their corresponding angles equal; consequently, the opposite sides are parallel; namely, the side AB parallel to CD, and BD parallel to AC; and, therefore, the figure is a parallelogram. COR. 3. Hence, also, if the opposite angles of a quadrilateral be equal, the opposite sides will likewise be equal and parallel. For all the angles of the figure being equal to four right angles (Cor. 2. Th. 26.), and the opposite angles being mutually equal, each pair of adjacent angles must be equal to two right angles; therefore, the opposite sides must be equal and parallel. PROP. XXVIII. THEOR. The opposite sides and angles of a parallelogram are equal to one another, and the diagonal bisects it; that is, divides it into two equal parts. Let ACDB be a parallelogram, of which BC is a diameter; the opposite sides and angles of the figure are equal to one another; and the diameter BC bisects it. Because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal (21. 1.) to one another; and because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal (21. 1.) to one another; wherefore the two triangles ABC, CBD have two angles ABC, BCA in one, equal to two angles BCD, CBD in the other, each to each, and the side BC, which is adjacent to these equal angles, common to the two triangles; therefore their other sides are equal, each to each, and the third angle of the one to the third angle of the other (Th. 2.); viz. the side AB to the side CD, and AC to BD, and the angle BAC equal to the angle BDC. And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, the whole angle ABD is equal to the whole angle ACD: And the angle BAC has been shewn to be equal to the angle BDC: therefore the opposite sides and angles of a parallelogram are equal to one another; also, its diameter bisects it; for AB being equal to CD, and BC common, the two AB, BC are equal to the two DC, CB, each to each; now the angle ABC is equal to the angle BCD; therefore the triangle ABC is equal (1. 1.) to the triangle BCD, and the diameter BC divides the parallelogram ACDB into two equal parts. COR. 1. equal. Two parallel lines, included between two other parallels, are COR. 2. If one angle of a parallelogram be a right angle, all the other three will also be right angles, and the parallelogram will be a rectangle. Hence, two parallels are every where equally distant. COR. 3. COR. 4. Hence, also, the sum of any two adjacent angles of a parallelogram is equal to two right angles. PROP. XXIX. THEOR. Parallelograms upon the some base and between the same parallels, are equivalent to one another. (SEE THE 2d And 3d figures.) Let the parallelograms ABCD, EBCF be upon the same base BC, and between the same parallels AF, BC; the parallelogram ABCD is equal to the parallelogram EBCF. If the sides AD, DF of the parallelograms ABCD, DBCF opposite to the base BC be terminated in the same point D; it is plain that each of the parallelograms is double (28. 1.) of the triangle BDC; and they are therefore equal to one another. D B C F But, if the sides AD, EF, opposite to the base BC of the parallelograms ABCD, EBCF, be not terminated in the same point; then, because ABCD is a parallelogram, AD is equal (28. 1.) to BC; for the same reason EF is equal to BC; wherefore AD is equal (1. Ax.) to EF; and DE is common; therefore the whole, or the remainder, AE is equal (2. or 3. Ax.) to the whole, or the remainder DF; now AB is also equal to DC; therefore the two EA, AB are equal to the two FD, DC, each to each; but the ex terior angle FDC is equal (21. 1.) to the interior EAB, wherefore the base EB is equal to the base FC, and the triangle EAB (1. 1.) to the triangle FDC. Take the triangle FDC from the trapezium ABCF, and from the same trapezium take the triangle EAB; the remainders will then be equal (3. Ax.), that is, the parallelogram ABCD is equal to the parallelogram EBCF. PROP. XXX. THEOR. Parallelograms upon equal bases, and between the same parallels, are equivalent to one another. Let ABCD, EFGH be parallelograms upon equal bases BC, FG, and between the same parallels AH, BG; the parallelogram ABCD is equal to EFGH. A B DE C F H G Join BE, CH; and because BC is equal to FG, and FG to (28. 1.) EH, BC is equal to EH; and, they are parallels, and joined towards the same parts by the straight lines BE, CH: But straight lines which join equal and parallel straight lines towards the same parts, are themselves equal and parallel (27. 1.); therefore, EB, CH are both equal and parallel, and EBCH is a parallelogram; and it is equal (29. 1) to ABCD, because it is upon the same base BC, and between the same parallels BC, AH; For the like reason, the parallelogram EFGH is equal to the same EBCH: Therefore also the parallelogram ABCD is equal to EFGH. PROP. XXXI. THEOR. Triangles upon the same base, and between the same parallels, are equivalent to one another. Let the triangles ABC, DBC be upon the same base BC, and between the same parallels, AD, BC: The triangle ABC is equal to the triangle DBC. Produce AD both ways to the points E, F, and through B draw BE parallel to CA; and through C draw CF parallel to BD: Therefore, each of the figures EBCA, DBCF is a parallelogram; and EBCA is equal (29. 1.) to DBCF, because they are upon the same base BC, and between the same pa rallels BC, EF; but the triangle ABC is the half of the parallelogram EBCA, because the diameter AB bisects (28. 1.) it; and the triangle DBC is the half of the parallelogram DBCF, because the diameter DC bisects it; and the halves of equal things are equal (7. Ax.); therefore the triangle ABC is equal to the triangle DBC. PROP. XXXII. THEOR. Triangles upon equal bases, and between the same parallels, are equivalent to one another. Let the triangles ABC, DEF be upon equal bases BC, EF, and between the same parallels BF, AD: The triangle ABC is equal to the triangle DEF. Produce AD both ways to the points G, H, and through B draw BG parallel to CA, and through F draw FH parallel to ED: G A D H Then each of the figures GBCA, DEFH is a parallelogram; and they are equal to (30. 1.) one another, because they are upon equal bases BC, EF, and between the same parallels BF, GH; and the triangle ABC is the half (28. 1.) of the parallelogram GBCA, because the diameter AB bisects it; and the triangle DEF is the half (28. 1.) of the parallelogram DEFH, because the diameter DF bisects it; But the halves of equal things are equal (7. Ax.); therefore the triangle ABC is equal to the triangle DEF. B CE PROP. XXXIII. THEOR. F Equivalent triangles upon the same base, and upon the same side of it, are between the same parallels. Let the equal triangles ABC, DBC be upon the same base BC, and upon the same side of it; they are between the same parallels. Join AD: AD is parallel to BC; for, if it is not, through the point A draw AE parallel to BC, and join EC: The triangle ABC, is equal (31. 1.) |