PROPOSITION XXXVII. THEOREM.-If triangles (ABC and DBC) are upon the same base and between the same parallels, they are equal to one another in area. CONSTRUCTION. Produce AD both ways to the points E and F; through B draw BE parallel to CA (a), and through C draw CF parallel to BD (a). DEMONSTRATION. Because the parallelograms EBCA and DBCF are upon the same base and between the same parallels, they are equal in area (b); and because the diagonals BA and CD bisect the equal parallelograms (c), the triangle ABC is half the parallelogram EBCA, and the (a) I. 31 (b) I. 35. (c) I. 34. (d) Ax. 7. triangle DBC is half the parallelogram DBCF; therefore the triangle ABC is equal in area to the triangle DBC (d). PROPOSITION XXXVIII. THEOREM.-If triangles (ABC and DEF) are upon equal bases and between the same parallels, they are equal to one another in area. CONSTRUCTION. Produce AD both ways to the points G and H; through B draw BG parallel to CA (a), and through F draw FH parallel to ED (a). G DEMONSTRATION. Because the parallelograms GBCA and DEFH are upon equal bases and between the same parallels, they are equal in area (b); and because the diagonals BA and FD bisect the equal parallelograms (c), the triangle ABC is half the parallelogram GBCA, and the triangle DEF is half the parallelogram DEFII; therefore the triangle ABC is equal to the triangle DEF (d). COROLLARY. Hence a straight line drawn from the vertex of a triangle bisecting its base, also bisects the triangle. PROPOSITION XXXIX. THEOREM.-If triangles equal in area (ABC and DBC) be upon the same base, and upon the same side of it, they are between the same parallels. DEMONSTRATION. Draw AD; then it is parallel to BC; for if not, through the point A draw AE parallel to BC (a), and join EC. Then the triangles ABC and EBC are equal because they are upon the same base (b) and between the same parallels (c); but ABC is also equal to DBC (6); therefore DBC is equal to EBC (d), the greater to the less; which is absurd. Therefore AE is not parallel to BC; and in the same manner it may be proved that no other line than AD is parallel to it; therefore AD is parallel to BC. I. 31. E (b) Hypoth. (c) I. 37. (d) Ax. 1. SCHOLIUM. This proposition is the converse of the thirty-seventh, and is demonstrated by the "reductio ad absurdum." PROPOSITION XL. THEOREM.-If triangles equal in area (ABC and DEF) are upon equal bases in the same straight line, and on the same side of it, they are between the same parallels. DEMONSTRATION. Draw AD; then it is parallel to BF; for if not, through the point A draw AG parallel to BF (a), and join GF. Then the triangles ABC and GEF are equal, because they are on equal bases (6), and between the same parallels (c); but ABC is also equal to DEF (b), therefore DEF is equal to GEF (d), the greater to the less; which is absurd. Therefore AG is not parallel to BF; and in the same manner it may be proved that no other line than AD is parallel to it; therefore AD is parallel to BF. B (a) I. 31. SCHOLIUM. This proposition is the converse of the thirty-eighth, and is demonstrated by the "reductio ad absurdum." The following corollaries may be drawn from this and the preceding proposition. COROLLARY 1. Any parallelogram or triangle is equal in area to a rightangled parallelogram or triangle having an equal base and altitude. COROLLARY 2. A straight line DE bisecting the two sides of a triangle is parallel to its base. For join BE and CD; then, because both the lines BE and CD bisect the triangle ABC (a), therefore the triangles DBC and EBC are equal in area; but they are also on the same base, and therefore between the same parallels DE and BC (b). COROLLARY 3. Straight lines joining the points of bisection ABC of the three sides of a triangle, divide it into four equal triangles. COROLLARY 4. The straight lines joining the points of bisection of the three sides of a triangle are respectively equal to half the parallel sides. I. 38, cor. COROLLARY 5. If the sides of a quadrilateral figure ABCD be bisected, and the points of bisection of each pair of conterminous sides joined by straight lines, those lines will form a parallelogram EFGH whose area is equal to half that of the quadrilateral. Draw AC and BD. The lines EH and FG are equal; being each equal to half AC (a), they are also parallel (b); therefore EFGH is a parallelogram (c). Further, the triangle DEH is equal in area to one-fourth of DAC, and BFG is equal in area to one-fourth of BAC (d); therefore the two triangles DEH and BFG are together equal in area to one-fourth of the whole figure ABCD; in like manner the two triangles AEF and CHG may be shown to be together equal to one-fourth of the whole; therefore DEH, AEF, BFG, and CHG are together equal in area to one-half the whole figure ABCD, and so the parallelogram EFGH must also be equal in area to half the same. be (a) Ax. 7, and I. 40, cor. 4. PROPOSITION XLI. THEOREM.-If a parallelogram (ABCD) and a triangle (EBC) upon the same base, and between the same parallels, the parallelogram is double of the triangle. CONSTRUCTION. Draw the diagonal AC. DEMONSTRATION. Then, because the two triangles ABC and EBC are on the same base and between the same parallels (a), therefore they are equal (b). But the parallelogram ABCD is double of the triangle ABC (c), therefore ABCD is also double of EBC. B C (a) Hypoth. COROLLARY. Hence, if a parallelogram and a triangle have equal bases and are between the same parallels, the parallelogram is double of the triangle. PROPOSITION XLII. PROBLEM. TO Construct a parallelogram equal in area to a given triangle (ABC) and having an angle equal to a given rectilineal angle (D). SOLUTION. Bisect BC in E (a), join AE, and at the point E make the angle FEC equal to the given angle D (b); also, through A draw AG parallel to BC (c), and through C draw CG parallel to EF (c). FECG will be the parallelogram required. (a) I. 10. (b) I. 23. I. 31. I. 38. (d) Const. (ƒ) I. 41. (g) Ax. 6. DEMONSTRATION. Because the triangles ABE and AEC are upon equal bases BE and EC, and between the same parallels BC and AG (d), therefore they are equal in area (e); and the triangle ABC is double of the triangle ABE; but, because the parallelogram FECG and the triangle AEC are upon the same base EC and between the same parallels BC and AG (d), therefore the parallelogram is double of the triangle (f); and the parallelogram FECG is equal in area to the triangle ABC (g), and has one of its angles equal to the given angle D. PROPOSITION XLIII. THEOREM. The complements (BK and KD) of the parallelograms (EH and GF), which are about the diagonal of any parallelogram (ABCD), are equal in area to one another. DEMONSTRATION. Because the diagonal of a parallelogram bisects it (a), the triangle ABC is equal to the triangle ADC, and the triangles AEK and KGC to the triangles AHK and KFC; then, if from the equals ABC and ADC the equals AEK and AHK, and also the equals KGC and KFC, be taken away, the remaining complements, BK and KD, will be equal in area (b). I. 34. Ax. 3. COROLLARY. The parallelograms about the diagonal, and also their complements, are equiangular with the original parallelogram. PROPOSITION XLIV. PROBLEM.-Upon a given finite straight line (AB) to construct a parallelogram which shall be equal in area to a given triangle (C), and have one of its angles equal to a given rectilineal angle (D). SOLUTION. Produce AB to E, and upon BE construct a parallelogram BEFG equal in area to the triangle C, and having the angle EBG equal to the given angle D (a). Produce FG to H, through A draw ÀÍ parallel to GB (b), and join HB. Then, because the straight line FH falls upon the two parallel lines FE and HA, the angles F and AHF are together equal to two right angles (c); and therefore the angles F and BHF are together less than two right angles; but the angles F and BHF are the interior angles made by HF with FE and HB on the same side; wherefore, if the straight lines FE and HB be produced, they shall meet (d). Let them meet in K, through K draw KL parallel to EA (b), and produce GB and HA to meet KL in the points M and L; then BALM is the parallelogram required. (a) I. 42. (b) I. 31. (c) I. 29. (d) Theor., I. 29. (e) I. 43. (f) Constr. (g) Ax. 1. DEMONSTRATION. Because FHLK is a parallelogram, of which HK is the diagonal, GA and EM the parallelograms about that diagonal, and FB and BL their complements, therefore BL is equal to FB (e); but FB is equal to the triangle C (ƒ), therefore BL is equal to the triangle C (g). And because the angle GBE is equal to the angle ABM (h), and also to the angle D (f), the angle ABM is equal to the angle D (g). Therefore the parallelogram BL constructed upon the given line AB is equal in area to the given triangle C, and has an angle equal to the given angle D. PROPOSITION XLV. PROBLEM.-TO construct a parallelogram equal in area to a given rectilineal figure (ABCD) and having an angle equal to a given rectilineal angle (E). SOLUTION. From any angle D draw a line DB dividing the given figure into triangles. Then construct a parallelogram FH equal in area to the triangle ABD, G L |