rallels BC, AE. But the triangle ABC is 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 other line but AD is parallel to BC; AD is therefore parallel to it.

COR.-This proposition is true of parallelograms.

For, if ABC, BDC, be the halves of the parallelograms, then, 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).

PROPOSITION XL. THEOREM.

Equal triangles upon the same side of equal bases, that are in the same straight line, are between the same parallels.

Let the equal triangles ABC, DEF, be upon the same side of equal bases BC, EF, in the same straight line BF; they are between the same parallels.

Join AD; AD is parallel to BC; for, if it is not, through A draw (I.31) AG parallel to BF, and join GF. The triangle ABC is equal to the triangle GEF

C

(1.38), 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; therefore also the triangle DEF is equal to the triangle GEF, the greater to the less, which is impossible; therefore 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; AD is therefore parallel to BF.

COR. This proposition is also true of parallelograms. It may be proved in the same manner as the Cor. to last proposition.

PROPOSITION A. THEOREM.

Equal triangles, between the same parallels, are upon equal bases.

Let the areas of the triangles ABC, DEF, be equal, and

let them lie between the same parallels AG, CF, then their bases AB, DE, are equal.

=

If AB be not equal to DE, let it be greater, and make DG AB, and join G, F. Then, since AB = DG, the triangles ABC, DGF, are equal (I. 38);

A

B D

EG

but the triangles ABC, DEF, are given equal; therefore the triangle DEF is equal to DGF, the less to the greater, which is impossible. Therefore the base AB is not greater than DE, and in a similar manner it may be proved not to be less; therefore AB=DE.

COR. This proposition is true of parallelograms.

For, if ABC, DEF, be the halves of the parallelograms, they are equal, and therefore AB=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.

If a parallelogram and a triangle be upon the same base, and between the same parallels, the parallelogram shall be double of the triangle.

Let the parallelogram ABCD and the triangle EBC be the same base BC, and between the same parallels BC,

upon

AE, the parallelogram ABCD is double of the triangle EBC.

Join AC; 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 diameter AC divides it into two equal parts; wherefore ABCD is also double of the triangle EBC.

PROPOSITION XLII. PROBLEM.

To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

Let ABC be the given triangle, and D the given rectilineal angle. It is required to describe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D.

F

E

Bisect (I. 10) BC in E, 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; therefore FECG is a parallelogram. And because BE is equal to EC, the triangle ABE is likewise equal to the B triangle AEC (I. 38), since they are upon equal bases BE, 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), because it is upon the same base, and between the same parallels; therefore the parallelogram FECG is equal to the triangle ABC, and it has one of its angles CEF equal to the given angle D; wherefore there has been described a parallelogram FECG equal to a given triangle ABC, having one of its angles CEF equal to the given angle D.

PROPOSITION XLIII. THEOREM.

The complements of the parallelograms which are about the diagonal of any parallelogram, are equal to one another. Let ABCD be a parallelogram, of which the diagonal is

E

AC; let EH, FG, be the parallelograms about AC; that is, through which AC passes, and BK, KD, the other parallelograms which make up the whole figure ABCD, which are therefore called the complements. The complement BK is equal to the complement KD.

Because ABCD is a parallelogram, and AC its diagonal, the triangle ABC is equal to the triangle ADC (Ï.34). And because EKHA 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 AĒK 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.

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.

Let AB be the given straight line, and C the given triangle, and D the given rectilineal angle. It is required to apply to the straight line AB a parallelogram equal to the triangle C, and having an angle equal to D.

E

Make (I. 42) the parallelogram BEFG equal to the triangle C, and having the angle EBG equal to the angle D, so 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, and join HB. Then because the straight line HF falls upon the parallels AH, EF, the

D

M

angles AHF, HFE, are together equal to two right angles (1.29); wherefore the angles BHF, HFE, are less than two right angles. But straight lines, which, with another straight line, make the interior angles upon the same side less than two right angles, do meet if produced far enough;

therefore HB, FE, shall meet, if produced; let them meet in K, and through K draw KL parallel to EA or FH, and produce HA, GB, to the points L, M. Then HLKF is a parallelogram, of which the diagonal is HK ; and AG, ME, are the parallelograms about HK; and LB, BF, are the complements; therefore LB is equal to BF (I. 43). But BF is equal to the triangle C; wherefore LB is equal to the triangle C; and because the angle GBE is equal to the angle ABM (I. 15), and likewise to the angle D; the angle ABM is equal to the angle D; therefore the parallelogram LB is applied to the straight line AB, is equal to the triangle C, and has the angle ABM equal to the angle D.

PROPOSITION XLV. PROBLEM.

To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle.

Let ABCD be the given rectilineal figure, and E the given rectilineal angle. It is required to describe a parallelogram equal to ABCD, and having an angle equal to E.

F

G

I.

c

Join DB, and describe (I. 42) the parallelogram FH equal to the triangle ABD, and having the angle HKF equal to the angle E; and (I. 44) to p the straight line GH apply the parallelogram GM equal to the triangle DBC, having the angle GHM equal to the angle E. And because the

E

B K H

angle E is equal to each of the angles FKH, GHM, the angle FKH is equal to GHM; add to each of these the angle KHG; therefore the angles FKH, KHG, are equal to the angles KHG, GHM; but FKH, KHG, are equal to two right angles (I. 29); therefore also KHG, GHM, are equal to two right angles; therefore KH is in the same straight line (I. 14) with HM. And because the straight line HG meets the parallels KM, FG, the alternate angles MHG, HGF, are equal; add to each of these the angle HĞL; therefore the angles MHG, HGL, are equal to the angles HGF, HGL; but the angles MHG, HGL, are equal to two right angles; wherefore also the angles HGF, HGL, are