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. B (a) I. 10. (b) I. 23. (c) I. 31. (d) Const. (e) I. 38. I. 41. 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). E (a) I. 34. (b) 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). (a) I. 42. B 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 AH 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. (c) I. 29. (e) I. 43. (f) Constr. g) Ax. 1. (h) I. 15. 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 (ƒ), 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, F G and having the angle K equal to the given angle E (a); and upon the straight line GH construct a parallelogram GM equal in area to the triangle DBC, and having the angle GIIM equal to the given angle E (b); then FKML will be the required parallelogram. DEMONSTRATION. Because the straight lines FK and GH are parallel, therefore the internal angles K and GHK are together equal to two right angles (c); but the angles K and GHM being both equal to the given angle E (d), are equal to one another (e); therefore the angles GHK and GHM are together equal to two right angles (e), and therefore KH and HM are in the same straight line (f). Because the straight line GH intersects the parallels FG and KM, the alternate angles FGH and GHM are equal (c); but the internal angles GHM and HGL are together equal to two right angles (c), therefore the angles FGH and HGL are together equal to two right angles (e), and therefore FG and GL are in the same straight line (f). Then, because FK and LM are both parallel to GH, therefore they are parallel to each other (9); and FL being paralleĺ to KM, FKML is a parallelogram. And because the parallelogram FH is equal in area to the triangle ABD, and the parallelogram GM to the triangle DBC, therefore the whole parallelogram FKML is equal in area to the whole figure ABCD (h), and has the angle K equal to the given angle E. COROLLARY. By means of this proposition, and that immediately preceding it, a parallelogram can be constructed on a given line equal in area to a given rectilineal figure, and having an angle equal to a given rectilineal angle, by constructing on the given line a parallelogram equal in area to the first triangle ABD. PROPOSITION XLVI. PROBLEM.-Upon a given finite straight line (AB) to construct a square. SOLUTION. From the point A draw AC perpendicular to AB (a), and make AD equal to AB (b); through the point D draw DE parallel to AB (c), and through the point B draw BE parallel to AD (c); then DABE is the required square. DEMONSTRATION. The side DE being parallel to the opposite side AB, and the side BE to ÂD (d), the figure DABE is a parallelogram (e); and therefore its opposite sides are equal (f), that is DE to AB, and BE to AD; and because AD is equal to AB (d), therefore all the sides are equal (g); but the angle A is a right angle, wherefore the four-sided figure DABE has all its sides equal and one of its angles a right angle; therefore it is a square (h), and it is constructed on the line AB. (d) Constr. SCHOLIUM. The definition of a square, as given by Euclid, viz. a four-sided figure, which has all its sides equal, and all its angles right angles," is more than sufficient, and really involves a theorem. It is only necessary to state in the definition that one of its angles is a right angle, and the proposition that its remaining angles are right may be demonstrated as follows. THEOREM. If a four-sided figure (DABE) be a square, all its angles are right angles. DEMONSTRATION. The opposite sides AB and DE are parallel (a), and they are met by the line AD, therefore the angles A and D are together equal to two right angles (b); but A is a right angle (a), therefore D is a right angle. And because, in the parallelogram DABE, the angles E and B are respectively opposite to A and D, which are right angles, therefore É and B are right angles; and therefore all the angles of the square DABE are right angles. COROLLARY 1. If two squares are constructed on equal straight lines AB and CD, they are equal. DEMONSTRATION. Draw the diagonals EB and GD. Because, in the triangles EAB and GCD, the sides EA and AB are respectively equal to GC and CD (a), and the angle A to the angle C, therefore the triangles are equal (b). And because the squares AF and CH are doubles of the triangles EAB and GCD (c), therefore they are equal (d). COROLLARY 2. If two squares AF and CH are equal. DEMONSTRATION. For, if it be possible, let one of them AF be the greater; take AK equal to CD, and AI equal to CG (a), and join IK. Then the triangles IAK and GCD are equal (b); but the triangles EAB and GCD are equal, being halves of the equal squares AF and CH (c), therefore the triangle IAK is equal to EAB (d), a part to the whole, which is absurd; therefore neither of the sides AB or CD is greater than the other, but they are equal. COROLLARY 3. To construct a rectangle under two given finite straight lines F and AB. SOLUTION. From the point A draw AC perpendicular to AB (a), and make AD equal to F(b); through the point D draw DE parallel to AB (c), and through the point B draw BE parallel to AD (c); then DABE is the rectangle required. E (a) Constr. (b) I. 29. (a) Hypoth. and Def. equal, their sides are D G H C (a) I. 3. Ax. 1. (a) I. 11. B DEMONSTRATION. The side DE being parallel to the opposite side AB, and the side BE to AD (d), the figure DABE is a parallelogram (e); and the angle A is a right angle (d), therefore it is a rectangle (f), and it is constructed under the two straight lines F and AB. d) Constr. Def. 27. PROPOSITION XLVII. THEOREM.-If a triangle (ABC) be right-angled, the square which is constructed upon the side (BC) subtending the right angle is equal in area to the sum of the squares constructed upon the sides (AB and AC) which form the right angle. CONSTRUCTION. On the sides AB, BC, and AC, construct the squares BG, (a) I. 46. DEMONSTRATION. Because the angles FBA and CBD are both right angles (a), therefore they are equal (c); add to both the angle ABC, and the angle FBC is equal to ABD (d). Because the sides FB and BC are respectively equal to AB and BD (e), and the angle FBC to the angle ABD, therefore the triangle FBC is equal to the triangle ABD (f). Because the angles GAB and BAC are both right angles, therefore GA and AC are in the same straight line (g). Now the parallelogram BL is double of the triangle ABD, because they are on the same base BD and between the same parallels BD and AL (h); and the square GB is double of the triangle FBC, being on the same base FB and between the same parallels FB and GC (h). But the doubles of equals are equal to one another (i), and therefore the parallelogram BL is equal in area to the square GB. And in the same manner, by joining AE and BK, it may be proved that the parallelogram CL is equal in area to the square CH. Therefore the whole square BDEC is equal in area to the two squares BG and CH. Def. 28. (ƒ) I. 4. (h) I. 41. (i) Ax. 6. |