to each of these equals add the angle ACB; and 4. The angles ACD, ACB, are equal to the three angles but the angles ACD, ACB, are equal (I. 13.) to two right angles; therefore also 5. The angles CBA, BAC, ACB, are equal to two right angles. Wherefore, if a side of any 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 the figure has sides. E D For any rectilineal figure ABCDE can be divided into as many triangles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles. And by the preceding proposition, ali 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; and the same angles are equal to the angles of the figure, together with the angles at the point F, which is the common vertex of the triangles; that is, (I. 15. Cor. 2.) together with four right angles. Therefore 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. Because every interior angle ABC, with its adjacent exterior ABD, is equal (I. 13.) 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, by the foregoing corollary, they are equal to all the interior angles of the figure, together with four right angles; therefore all the exterior angles are equal to four right angles. PROP. XXXIII.-THEOREM. 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. B Join BC; and because AB is parallel to CD, and BC meets them, (I. 29.) 1. The alternate angles ABC, BCD, are equal. 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 the angle BCD; therefore (I. 4.) 2. 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 3. 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, (I. 27.) 4. AC is parallel to BD; and it was shown to be equal to it. Therefore, straight lines, &c. Q.E.D. PROP. XXXIV.-THEOREM. 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. A B Because AB is parallel to CD, and BC meets them, (I. 29.) 1. The alternate angles ABC, BCD, are equal to one and because AC is parallel to BD, and BC meets them, (1. 29.) wherefore the two triangles ABC, CBD, have two angles ABC, BCA, in and because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB; 4. 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, their 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 has been proved equal to the angle BCD; therefore (I. 4.) 5. The triangle ABC is equal to the triangle BCD, and the diameter BC divides the parallelogram 4CDB into two equal parts. Q.E.D. 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. B B E A E D F B 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 (I. 34.) 1. Each of the parallelograms ABCD, DBCF, is double of the triangle BDC; and therefore (Ax. 6.) 2. The parallelograms ABCD, DBCF, are equal to one another. 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 (I. 34.) to BC; for the same reason, EF is equal to BC; wherefore (Ax. 1.) 1. AD is equal to EF; and DE is common; therefore (Ax. 2. or 3.) 2. The whole, or the remainder, AE is equal to the whole, or the remainder, DF; AB also is equal (I. 34.) to DC; and the two EA, AB, are therefore equal to the two FD, DC, each to each; and the exterior angle FDC is equal (I. 29.) to the interior EAB, therefore (I. 4.) 3. The base EB is equal to the base FC, and the triangle EAB equal to the triangle FDC. Take the triangle FDC from the trapezium ABCF, and from the same trapezium take the triangle EAB; the remainders therefore are equal (Ax. 3.), that is, 4. The parallelogram ABCD is equal to the parallelogram EBCF Therefore parallelograms upon the same base, &c. Q.E.D. PROP. XXXVI.-THEOREM. Parallelograms upon equal bases and between the same parallels, are equal 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. DE Join BE, CH; and because BC is equal to FG, and FG (I. 34.) to EH, 1. BC is equal to EH; and they are (Hyp.) 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 (I. 33.); therefore 2. EB, CH, are both equal and parallel, because they are upon the same base BC, and between the same parallels BC, AH. For the like reason 5. The parallelogram EFGH is equal to the same EBCH. Therefore also 6. The parallelogram ABCD is equal to EFGH. Wherefore, parallelograms, &c. Q.E.D. PROP. XXXVII.—THEOREM. Triangles upon the same base and between the same parallels, are equal 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. E Therefore Produce AD both ways to the points E, F, and through B draw (I. 31.) BE parallel to CA; and through C draw CF parallel to BD. (I. 34. Def.) 1. and (I. 35.) Each of the figures EBCA, DBCF, is a parallelogram; 2. EBCA is equal to DBCF, because they are upon the same base BC, and between the same parallels BC, EF; and (I. 34.) 3. The triangle ABC is the half of the parallelogram EBCA, because the diameter AB bisects it; and 4. The triangle DBC is the half of the parallelogram DBCF, because the diameter DC bisects it. But the halves of equal things are equal (Ax. 7.); therefore 5. The triangle ABC is equal to the triangle DBC. Wherefore, triangles, &c. Q.E.D. |