therefore also the angles BAC, ACB, CBA, are equal to two right angles. Wherefore, If a side of a triangle &c. Q. E. D. COR. 1. All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as E the figure has sides. D For any rectilineal figure ABCDE can be divided into as many triangles as the figure has sides, by drawing straight lines from any point F within the figure to each of its angles. Now, by the preceding proposition, all the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as there are sides of the figure: But all the angles of the triangles make up the angles of the figure, together with the angles at the point F, that is, together with four right angles (1. 15, Cor. 2): Wherefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides. COR. 2. All the exterior angles of any rectilineal figure are together equal to four right angles. A DB For every interior angle ABC, with its adjacent exterior ABD, is equal to two right angles (1. 13); therefore all the interior angles, together with all the exterior, are equal to twice as many right angles as there are sides of the figure, that is, by the foregoing corollary, are equal to all the interior angles, together with four right angles: Wherefore all the exterior angles are equal to four right angles. PROP. XXXIII. THEOR. 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, joined towards the same parts by the straight lines AC, BD: AC, BD, are themselves equal and parallel. Join BC: Then because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal (1. 29): And because AB is equal to CD (Hyp.), and BC common to the two triangles ABC, B DCB, the two sides AB, BC, are equal to the two DC, CB, and the angle ABC is equal to the angle BCD-therefore the base AC is equal to the base BD, and the triangle ABC to the triangle BCD, and the other angles to the other angles, 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 falling on the two straight lines AC, BD, makes the alternate angles ACB, CBD equal to one another, AC is parallel to BD (1. 27); and it has been shewn to be equal to it. Wherefore, Straight lines &c. Q. E.D. PROP. XXXIV. THEOR. The opposite sides and angles of parallelograms are equal to one another, and the diameter bisects them, that is, divides them into two equal parts. N.B.-A parallelogram is a four-sided figure, of which the opposite sides are parallel; and the diameter is the straight line joining two of its opposite angles. 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 A C B D to one another (1. 29); and because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD, are equal to one another: Therefore the two triangles ABC, CBD have two angles ABC, BCA in the one, equal to two angles BCD, CBD in the other, each to each, and one side BC common to the two triangles, which is adjacent to their equal angles-therefore (1.26) their other sides are equal, each to each, and the third angle of the one to the third angle of the other viz. the side AB to the side CD, and AC to BD, and the angle BAC to the angle BDC: Then because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, therefore the whole angle ABD is equal to the whole angle ACD; and the angle BAC has been shown to be equal to the angle BDC; therefore the opposite sides and angles of parallelograms are equal to one another: Also, the diameter bisects them; for AB being equal to CD, and BC common, the two AB, BC are equal to the two DC, CB, each to each, and the angle ABC is equal to the angle BCDtherefore the triangle ABC is equal to the triangle BCD, and the diameter BC divides the parallelogram ACDB into two equal parts. Wherefore, The opposite sides &c. Q. E. D. PROP. XXXV. THEOR Parallelograms upon the same base, and between the same parallels, are equal to one another. Let the parallelograms ABCD, EBCF be upon the same base BC, and between the same parallels AF, BC; the parallelogram ABCD shall be equal to the parallelogram EBCF. If (fig. 1) the points D, E, coincide, it is plain that each of the parallelograms ABCD, DBCA, is double of the triangle BDC (1. 34), and they are therefore equal to one another. But if (fig. 2 and 3) the points D, E, do not coincide, then, because ABCD is a parallelogram, AD is equal to BC (1.34): For the like reason EF is equal to BC; and therefore AD is equal to EF: In fig. 2 add DE to each, in fig. 3 take DE from each; then the whole, or the remainder AE, is equal to the whole, or the remainder DF: And AB is equal to DC; therefore the two EA, AB are equal to the two FD, DC, each to each, and the angle EAB is equal to the angle FDC (1. 29)—therefore the base EB is equal to the base FC, and the triangle EAB to the triangle FDC: Take the triangle FDC from the trapezium ABCF, and from the same trapezium take the triangle EAB; then the remainders are equal, that is, the parallelogram ABCD is equal to the parallelogram EBCF. Wherefore, Parallelograms upon the same base &c. Q. E.D. PROP. XXXVI. THEOR. Parallelograms upon equal bases, and between the same parallels, are equal to one another. A DE H Let ABCD, EFGH be parallelograms upon equal bases BC, FG, and between the same parallels AH, BG: the parallelogram ABCD shall be equal to EFGH. Join BE, CH: Then because BC is equal to FG, and FG to EH (1. 34), BC is equal to EH: And they are also parallels, and joined towards the same parts by the straight lines BE, CH: But straight B C F lines which join the extremities of two equal and parallel straight lines towards the same parts, are themselves equal and parallel (1. 33); therefore EB, CH are both equal and parallel, and EBCH is parallelogram, and it is G equal to ABCD, because it is upon the same base BC, and between the same parallels BC, AH (1.35): For the like reason, the parallelogram EFGH is equal to the same EBCH: Therefore also the parallelogram ABCD is equal to EFGH. Wherefore, Parallelograms &c. Q.E. D. PROP. XXXVII. THEOR. Triangles upon the same base, and between the same Let the triangles ABC, DBC be upon the same base BC, and between the same parallels AD, BC: the triangle ABC shall be equal to the triangle DBC. Produce AD both ways to the points E, F, and through B draw BE parallel to CA (1. 31), and through C draw CF pa- E A B C F because they are upon the same base BC, and between the same parallels BC, EF (1.35): But the triangle E |