1 Because AB is parallel to CD, and BC meets them, (I. 29.) 1. The alternate angles. ABC, BCD, are equal to one another;, and because AC is parallel to BD, and BC meets them, (I. 29.) 2. The alternate angles ACB, CBD, are equal 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 one side BC common to the two triangles, which is adjacent to their equal angles; therefore their other sides shall be equal, each to each, and the third angle of the one to the third angle of the other (I. 26.), viz. 3. The side AB is equal to the side CD, and AC to BD, and the angle BAC is equal to the angle BDC; 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 XB 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 ACDB into two equal parts. Q.E.D. PROP. XXXV. -THEOREM. 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 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 4D, 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. l.) 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. А 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, and (I. 34. Def.) 3. EBCH is a parallelogram; and (I. 35.) 4. EBEH is equal to ABCD, 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. 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 en the same parallels AD, BC. The triangle ABC is equal to the triangle DBC. E Ꭺ ' Ꭰ 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. Therefore (I. 34. Def.) 1. Each of the figures EBCA, DBCF, is a parallelogram; and (I. 35.) 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. PROP. XXXVIII.-THEOREM. Triangles upon equal bases and between the same parallels, are equal to one another. Let the triangles ABC, DEF, be upon equal bases BC, EF, and between the same carallels BF, AD. The triangle ABC is equal to the i angle DEF. D Produce 4D both ways to the points G, H, and through B draw BG parallel (I. 31.) to CA, and through F draw FH parallel to ED. Then (I. 34. Def.) 1. Each of the figures GBCA, DEFH, is a parallelogram; and (I. 36.) 2. GBCA, DEFH, are equal to one another, because they are upon equal bases BC, EF, and between the same parallels BF, GH; and (I. 34.) 3. The triangle ABC is the half of the parallelogram GBCA, because the diameter AB bisects it; and 4. The triangle DEF is the half of the parallelogram DEH, because the diameter DF bisects it. But the halves of equal things are equal (Ax. 7.); therefore 5. The triangle ABC is equal to the triangle DEF. Wherefore, triangles, &c. Q.E.D. PROP. XXXIX. -THEOREM. Equal 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. B Join AD. AD is parallel to BC. For, if it is not, through the point A draw (I. 31.) AE parallel to BC, and join EC. Then (I. 37.) 1. The triangle ABC is equal to the triangle EBC, because they are upon the same base BC, and between the same parallels BC, AE. But the triangle ABC is equal to the triangle BDC (Hyp.); therefore also 2. The triangle BDC is equal to the triangle EBC, the greater to the less, which is impossible. Therefore 3. AE is not parallel to BC. In the same manner, it can be demonstrated that no other line but AD is parallel to BC; therefore 4. AD is parallel to BC. Wherefore, equal triangles upon, &c. Q.E.D. PROP. XL.-THEOREM. Equal triangles upon equal bases in the same straight line, and towards the same parts, are between the same parallels. Let the equal triangles ABC, DEF, be upon equal bases BC, EF, in , the same straight line BF, and towards the same parts; they are between the same parallels. A Join AD. AD is parallel to BC. For, if it is not, through A draw (I. 31.) 4G parallel to BF, and join GF. Then (I. 38.) AG 1. The triangle ABC is equal to the triangie GEF, because they are upon equal bases BC, EF, and between the same parallels BF, AG. But the triangle ABC is equal to the triangle DEF (Hyp.); therefore also 2. The triangle DEF is equal to the triangle GEF, the greater to the less, which is impossible. Therefore 3. AG is not parallel to BF. And in the same manner it can be demonstrated that there is no other parallel to it but AD; therefore 4. AD is parallel to BF. Wherefore, equal triangles, &c. Q.E.D. PROP. XLI.-THEOREM. If a parallelogram and a triangle be upon the same base, and between the same parallels, the parallelogram shall be double of the triangle. |