whence the triangles DCF, ABE, having also DC, CF, severally equal to AB, BE [34. 1], are equal [4. 1]; but the triangle DCF being taken from the quadrangle ABCF, the remains the parallelogram BD; and the triangle ABE being taken from the same quadrangle, there remains the parallelogram BF; the parallelograms BD, BF are therefore equal [Ax. 3].

Scholium.—The demonstration is evidently the same, though the point E should coincide with D, or be between the points A and D.

PROP. XXXVI. THEOR.

Parallelograms (BD, FH), on equal bases (BC, FG), and between the same parallels (BG, AH), are equal.

Join BE, CH, and because BC is equal to FG [Hyp.], and EH to FG (34. 1), BC is equal to EH Ax. 1); and BC, EH are parallel [Hyp.], therefore BE, CH, which join them, are parallel (33. 1), and BH is a parallelogram (Def. 35), and is equal to the parallelogram BD, being on the same base BC, and between the same parallels BC, AH (35. 1); and the parallelogram BH is also equal to FH, being on the same base EH, and between the same parallels BG, EH (35. 1); whence, the parallelograms BD and FH, being each equal to the parallelogram BH, are equal to each other.

PROP. XXXVII. THEOR.

Triangles (ABC, DBC), on the same base (BC), and between the same parallels (BC, AD), are equal.

Through B, draw BE parallel to CD, and through C, CF parallel to BA [31. 1]; E meeting AD produced in E and F; the parallelograms EC, BF are equal, being on the same base BC, and between the same parallels BC, EF [35. 1]; but the triangles DBC, ABC are the halves of these

parallelograms, because the diagonals BD, AC bisect them [34. 1]; therefore these triangles are equal [Ax. 7].

PROP. XXXVIII. THEOR.

Triangles (ABC, DEF), on equal bases (BC, EF), and between the same parallels (BF, AD), are equal.

Through B, draw BG parallel to CA, and through F, FH parallel to ED [31. 1], meeting AD produced in G and H; the parallelograms GC, EH are equal, being on equal bases BC, EF, and between the same parallels BF, GH, [36. 1]; whence, the triangles ABC, DEF, being the halves of these parallelograms [34. 1], are also equal [Ax. 7].

PROP. XXXIX. THEOR.

Equal triangles (ABC, DBC), on the same base (BC), and on the same side of it, are between the same parallels.

Join AD, which is parallel to BC; for, A if not, through A, draw AE parallel to BC [31. 1], meeting either side BD of the triangle DC, in a point E, different from the vertex, and join EC.

B

The triangles ABC, EBC, being on the same base BC, and between the same parallels BC, AE, are equal [37. 1]; but the triangle DBC is equal to ABC [Hyp.]; therefore the triangles EBC, DBC are equal [Ax. 1], part and whole, which is absurd; therefore AE is not parallel to BC. In like manner it may be shewn, that no other right line drawn through A. except AD, is parallel to BC; therefore AD is parallel to BC.

PROP. XL. THEOR.

Equal triangles (ABC, DEF), on equal bases in the same right line (BC, EF), and towards the same part, are between the same parallels.

D

B

Join AD which is parallel to BF; for, if not, through A, draw AG parallel to BF [31. 1], cutting either side. DE of the triangle DEF, in a point G different from the vertex, and join GF.

The triangles ABC, GEF, being on equal bases BC, EF, and between the same parallels BF, AG, are equal [38. 1]; but the triangle DEF is equal to ABC [Hyp.]; therefore the triangles DEF, GEF are equal [Ax. 1], whole and part, which is absurd; therefore AG is not parallel to BF. In like manner it may be shewn, that no other right line drawn through A, except AD, is parallel to BF; therefore AD is parallel to BF.

PROP. XLI. THEOR.

If a parallelogram (BD), and triangle (BEC), be on the same base (BC), and between the same parallels (BC, AE), the parallelogram is double to the triangle.

Join CA. The triangle BEC is A equal to BAC, being on the same base BC, and between the same parallels BC, AE, [37. 1]; but the parallelogram BD, being bisected by the diagonal AC [34. 1], is double the triangle BAC;

D

therefore the parallelogram BD is also double the triangle BEC.

PROP. XLII. PROB.

To constitute a parallelogram, equal to a given triangle (ABC), and having an angle, equal to a given rectilineal angle (D).

Through A, draw AG parallel to BC [31. 1], bisect BC in E (10. 1), and at the point E, with the right line EC, make the angle CEF equal to the given angle D (23. 1); through C, draw CG parallel to EF (31. 1), meeting AG in G, and join AE; ECGF is a parallelogram (Def. 35), and double the triangle AEC (41. 1); and since the triangles ABE, AEC are equal, being on equal bases BE. EC, and between the same parallels BC, AG (38. 1), the triangle ABC is double the triangle AEC; whence, the parallelogram EG and triangle ABC, being each of them double the triangle AEC, are equal (Ax. 6); and one angle FEC of the parallelogram is equal to the given angle D (Constr.). There is therefore constituted a parallelogram EG, equal to the given triangle ABC, and having an angle FEC, equal to the given angle D, as was required.

PROP. XLIII. THEOR.

In any parallelogram (BD), the complements (BK, KD), of the parallelograms (EH, GF), which are about the diagonal (AC), are equal.

For, because AC bisects the parallelogram BD (34. 1), the triangles ABC, ADC are equal; and, because the same diagonal bisects the parallelograms EH, GF (34. 1), the triangles AEK, KGC are severally equal to the triangles AHK, KFC; therefore the triangles AEK, KGC together, are equal to AHK, KFC together (Ax. 2), which being taken from the equal triangles ABC, ADC, the remaining complements BK, KD are equal (Ax. 3).

PROP. XLIV. PROB.

To a given right line (AB), to apply a parallelogram, equal to a given triangle (C), and having an angle, equal to a given rectilineal angle (D).

Make the parallelogram EFGH equal to C, having an angle EFG equal to D (42. 1); in EF produced, take F1 equal to AB (3. 1), complete the parallelogram GI (Cor. 6. 34. 1), and join FK; because the angle FGK is equal to the internal remote angle EHG (29. 1), and the angles FGK, FKG are together less than two right angles (17. 1), the angles EHK, FKH are also together less than two right angles, therefore HE, KF may be so produced towards E, F, as to meet (Theor. at 29. 1); let them, being produced, meet, as in L, and through L, draw LN parallel to EI, meeting GF, KI produced in M and N; make the angle BAO equal to IFM (23. 1), AO to FM, and complete the parallelogram AOPB (Cor. 6. 34. 1)