Given two parallelograms ABCD and EFGH upon equal bases BC and FG, and between the same parallels AH and BG; to prove that the parallelogram ABCD is equal to EFGH. (Const.) Join BE and CH; to FG, and FG to EH (I. 34), fore BC is equal to EH ; (Dem.) and because BC is equal there and they are parallels, and joined towards the same parts by the straight lines BE and CH; therefore (I. 33) EB and CH are both equal and parallel, is a parallelogram; wherefore EBCH and it is equal to B H ABCD (I. 35), because it is upon the same base BC, and between the same parallels BC and AH. For a like reason, the parallelogram EFGH is equal to the same parallelogram EBCH; therefore (Ax. 1) the parallelogram ABCD is equal to EFGH. PROPOSITION XXXVII. THEOREM. Triangles upon the same base, and between the same parallels, are equal to one another. Given the triangles ABC and DBC upon the same base BC, and between the same parallels AD and BC; to prove that the triangle ABC is equal to the triangle DBC. (Dem.) (Const.) Join AD, and produce it both ways to the points E and F, and through B draw BE parallel to CA (L. 31); and through C draw CF parallel to BD. Therefore, each of the figures EBCA and DBCF is a parallelogram; EBCA is equal to DBCF (I. 35), because they are upon the same base BC, and between the same parallels BC and EF; and E B A D F and the triangle ABC is the half of the parallelogram EBCA, because the diagonal AB bisects it (I. 34); and the triangle DBC is the half of the parallelogram DBCF, because the diagonal DC bisects it. But the halves of equal things are equal (Ax. 7); there. fore the triangle ABC is equal to the triangle DBC. Triangles upon equal bases, and between the same parallels, are equal to one another. Given the triangles ABC and DEF upon equal bases BC and EF, and between the same parallels BF, AD; to prove that the triangle ABC is equal to the triangle DEF. (Const.) Produce AD both ways to the points G and H, through B draw BG parallel (I. 31) to CA, and and through F draw FH parallel to ED. (Dem.) Then each of the figures and they are equal to GBCA and DEFH is a parallelogram; G A D H one another (I. 36), because they are upon equal bases BC, EF, and between the same parallels BF, GH; and the triangle ABC is the half of the parallelogram GBCA (I. 34), because the diagonal AB bisects it; and the triangle DEF is the half of the parallelogram DEFH, because the diagonal DF bisects it. But the halves of equal things are equal (Ax. 7); ABC is equal to the triangle DEF. B CE F therefore the triangle Equal triangles upon the same base, and upon the same side of it, are between the same parallels. Given the equal triangles ABC and DBC, upon the same base BC, and upon the same side of it; to prove that between the same parallels. they are for, if it is (Const.) Join AD; AD is parallel to BC; not, through the point A draw (I. 31) AE parallel to BC, and join EC. (Dem.) The triangle ABC is equal to the triangle EBC (I. 37), because it is upon the same base BC, and between the same parallels BC, AE. But the triangle ABC is given equal to the triangle BDC; therefore also, the triangle BDC is equal to the triangle EBC, the greater to the less, which is impossible; therefore AE is not parallel to BC. In the same manner, it can be demonstrated that no A B D E line, passing through A, can be parallel to BC except AD, which therefore is parallel to it. COR.-This proposition is true of parallelograms. then, For, if ABC, BDC be the halves of the parallelograms, since AD is parallel to BC, the sides of the two parallelograms passing through A and D, and being both parallel to BC, must lie in the same line with AD (Ax. 11). Equal triangles upon the same side of equal bases, that are in the same straight line, are between the same parallels. Given the equal triangles ABC and DEF upon the same side of equal bases BC and EF, in the same straight line BF; to prove that they are between the same parallels. (Const.) Join AD; AD is parallel to BC; for if it is not, through A draw (I. 31) AG parallel to BF, and join GF. (Dem.) The triangle ABC is equal to the triangle GEF because they are upon equal (I. 38), bases BC and EF, and between the same parallels BF, AG; but the triangle ABC is equal to the triangle DEF; therefore also the triangle DEF is equal to the triangle GEF, A D the greater to there the less, which is impossible; fore AG is not parallel to BF. In the same manner, it can be demonstrated that no line, passing through A, can be parallel to BC except AD, which therefore is parallel to it. COR.-This proposition is also true of parallelograms. It may be proved in the same manner as the Cor. to last proposition. Equal triangles, between the same parallels, are upon equal bases. Given the areas of the triangles ABC and DEF, equal to each other, and that they lie between the same parallels AG, CF; to prove that their bases AB and DE are equal. (Const.) If AB be not equal to DE, and make DG equal AB, and join G, F. (Dem.) Then, since AB is equal to DG, the triangles ABC and DGF are equal (I. 38); but the triangles ABC and DEF are given equal; therefore the triangle DEF is equal to DGF, the greater to the less, which is impossible. Therefore the base DE is not greater than AB, that it is not less; let DE be the greater, A F BD GE and in a similar manner it may be proved therefore AB is equal to DE. COR.-This proposition is true of parallelograms. For, if ABC and DEF be the halves of the parallelograms, they are equal, and therefore AB is equal to DE. Scholium 1.-The preceding seven propositions, with the corollaries to the last three, are also true, when the term altitude is substituted for the expression between the same parallels; observing, that in the last three propositions it is not necessary that the two triangles or parallelograms be upon the same sides of their bases, or that their bases be in the same straight line; for when the two figures do not fulfil these conditions, other two equal to them in every respect may be constructed so as to fulfil them (I. 22), and the demonstrations will apply to the latter (I. 34, Cor. 2). Schol. 2.-It appears from these propositions also, that of these three conditions-the equality of the bases, of the altitudes, and of the areas of two parallelograms or triangles-if any two be given, the third will also be fulfilled; that is, if the bases and altitudes be equal, the areas are equal; if the bases and areas be equal, the altitudes are equal; and if the areas and altitudes be equal, the bases are equal. PROPOSITION 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: Given the parallelogram ABCD and the triangle EBC upon the same base BC, and between the same parallels BC, AE; to prove that the parallelogram ABCD is double of the triangle EBC. (Const.) Join AC. (Dem.) Then the triangle ABC is equal to the triangle EBC (I. 37), because they are upon the same base BC, and between the same parallels BC, AE. But the parallelogram ABCD is double of the triangle ABC (I. 34), because the diagonal AC divides it into two equal parts; is also double of the triangle EBC. wherefore ABCD To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. A F G It is Given the triangle ABC and the rectilineal angle D. required to describe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D. (Const.) Bisect (I. 10) BC in E, and join AE, and at the point E in the straight line EC make (I. 23) the angle CEF equal to D; and through A draw AG parallel to EC (I. 31), and through C draw CG parallel to EF; then FECG is a parallelogram. (Dem.) And because BE is equal to EC, the triangle ABE is likewise equal to the triangle AEC (I. 38), since they are upon equal bases BE and EC, and between the same parallels BC, AG; therefore the triangle ABC is double of the triangle AEC. And the parallelogram FECG is likewise double of the triangle AEC (I. 41), upon the same base and between the same parallels; therefore the parallelogram FECG is equal to the triangle ABC (Ax. 6), and it has one of its angles CEF equal to the given B E C because it is angle D; wherefore there has been described a parallelogram FECG equal to the given triangle ABC, and having one of its angles CEF equal to the given angle D. The complements of the parallelograms which are about the diagonal of any parallelogram are equal to one another. And A H K D Given a parallelogram ABCD, of which the diagonal is AC; and let EH and FG be the parallelograms about AC, that is, through which AC passes; and BK and KD the other parallelograms which make up the whole figure ABCD, which are therefore called the complements; to prove that the complement BK is equal to the complement KD. (Dem.) Because ABCD is a parallelogram, E and AC its diagonal, the triangle ABC is equal to the triangle ADC (I. 34). because AEKH is a parallelogram, the diagonal of which is AK, the triangle AEK is equal to the triangle AHK. For the same reason, the triangle KGC is equal to the triangle KFC. Then, because the triangle AEK is equal to the triangle AHK, and the triangle KGC to KFC; the triangle AEK, together with the triangle KGC, is equal to the triangle AHK, together with the triangle KFC. But the whole triangle ABC is equal to the whole ADC; therefore the remaining complement BK is equal to the remaining complement KD (Ax. 3). Б Ꮐ PROPOSITION XLIV. PROBLEM. To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Given the straight line AB, the triangle C, and the rectilineal angle D. It is required to apply to the straight line AB a parallelogram equal to the triangle C, and having an angle equal to D. (Const.) Make (I. 42) the paral lelogram BEFG equal to the triand having the angle angle C, EBG equal to the angle D, C that BE be in the same straight line with AB, and produce FG to H; and through A draw (I. 31) AH parallel to BG or EF, because the straight line HF falls upon the parallels AH, EF, the angles AHF and HFE, are together equal to two right |