BC, and between the

same parallels AF, BC. The parallelogram AC is equal to the parallelogram BF.

Because each of the parallelograms AC, BF, is double (P. 11) of the triangle BDC. Therefore the parallelogram AC is equal (Ax. 4)to the parallelogram BF.

Next, let the sides AD, EF, opposite to the base BC, be not terminated in the same point.

Because AC is a parallelogram, AD is equal (P. 11) tỏ BC. For a similar reason, EF is equal to BC. Therefore AD is equal (Ax. 1) to EF; and DE is common to both Wherefore the whole, or the remainder AE, is equal to the whole, or the remainder DF (Ax. 2 or 3); and AB is equal (P. 11) to DC. Because in the triangles EAB, FDC, the sides AB, AE are equal to the sides DF, DC each to each; and the exterior angle FDC is equal (I. 8) to the interior and opposite angle EAB. Therefore the base FC is equal (P. 3) to the base EB, and the triangle FDC to the triangle EAB. From the figure ABCF, take the triangle FDC, and the remainder is the parallelogram AC. From the same figure take the triangle EAB and the remainder is the parallelogram BF. But when equals are taken from equals, or from the same, the remainders (Ax. 3) are equal. Therefore the parallelogram AC is equal to the parallelogram BF. Therefore, parallelograms upon the same &c. Q. E. D.

PROP. XIII. THEOREM. (E. I. 41.)

If a parallelogram and a triangle be upon the between the same parallels, the parallelogram is triangle.

same base and double of the

First, let ABCD be a parallelogram and ABC a tri

ABCD is double of the triangle ABC.

Next, let DBCF be a parallelogram and ABC a tri angle. Draw BG parallel to AC (P. 9). Produce FA to G.

Because GA is parallel to BC and AC to GB. Therefore GBCA is a parallelogram and AB being the diagonal, the parallelogram is double of the triangle ABC (P. 11). But the parallelogram BF is equal to the parallelogram GBCA (P. 12), and is therefore double of ABC. Wherefore if a parallelogram and a triangle &c. Q. E. D.

PROP. XIV. THEOREM. (E. I. 36.)

Parallelograms upon equal bases and between the same parallels are equal.

Let ABDC and EFGH be two parallelograms upon

angles BAE and EBH, the side AB is equal to the side

EH (P. 11) and EB is common. And because the angle ABE is equal to the angle BEH (P. 7). Therefore the two triangles are equal (P. 3).

But the parallelogram AD is double of the triangle ABE (P. 13). For the same reason the parallelogram EG is also double of the triangle BEH. Therefore the parallelograms AD and EG are equal (Ax. 4).

PROP. XV. THEOREM. (E. 1. 37, 38, 39, 40.)

Triangles upon equal bases and between the same parallels are equal; and conversely equal triangles upon equal bases and between the same straight lines are between the same parallels. Let ABC and DEF be two

triangles upon equal bases BC and EF. Draw CP parallel to AB and FS parallel to ED.

NN

Because BP is double of the triangle ABC (P. 13) and ES double of the triangle DEF (P. 13). And because BP and ES are equal (P. 14.). Therefore the triangles ABC and DEF are equal (Vx. 5).

Next, if the triangles be upon the same base (Def. 26) as ABC and DBC, then draw straight lines parallel to AC and BD and demonstrate the proposition by P. 12.

Conversely, let the triangles ABC and DEF or ABC and DBC be upon equal bases and between the same straight

lines AD, BF or

AD,BC. The triangles are between the same parallels, that is, AD and BF or AD and BC are parallels. Because

it can be proved that no other line than AD can be parallel to BF or BC. For, if possible, let AG be parallel to BF or BC. Join GF or GC.

Because, triangles upon equal bases and between the same parallels are equal (Ist case). Therefore ABC and GEF or ABC and GBC are equal; which is impossible (hyp). Therefore AG is not parallel to BF or BC but AD is parallel to it. Wherefore triangles, &c. Q.E.D.

Cor. Different triangles upon the same straight line, having their vertices at the same point, are between the same parallels.

PROP. XVI. THEOREM. (E. 6. 1.)

Triangles and parallelograms between the same parallels are to one another as their bases.

Let the triangles ABC and ACD be between the same parallels A and BD (15. cor). The base BC is to the base CD, so is the triangle ABC to the triangle ACD.

G. B C

Produce BD both ways to H and L, making CB, BG and GH equal and also CD, DK and KL equal.

Because CB, BG and GH are equal bases and between the same parallels A and HL. Therefore the triangles ABC, ABG and AGH are all equal (P. 15). Therefore whatever multiple the base HC is of the base BC, the same multiple is the triangle AHC of the triangle ABC. For the same reason, whatever multiple the base LC is of the base CD, the same multiple is the triangle ALC of the triangle ADC. But if the base HC be equal to the base : CL, the triangle AHC is also equal to the triangle ALC (P. 15); and if the base HC be greater than the base CL,

the triangle AHC is likewise greater than the triangle ALC and if less, less. Because there are four magnitudes, viz, the two bases BC and CD, and the two triangles ABC and ACD, and of the base BC, and the triangle ABC; the first and the third, any equimultiples whatever have been taken, viz., the base HC and the triangle AHC; and of the base CD, and the triangle ACD, the second and the fourth, any equimultiples whatever have been taken, viz., the base CL and the triangle ALC, and it has been shown, that if the base HC be greater than the base CL, the triangle AHC is greater than the triangle ALC; if equal, equal; and if less, less. Therefore, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD (Def. 30). Next, let the parallelograms ABCT

and TCDE be between the same parallels AE and BD. BC is to CD, so is the parallelogram ABCT to the parallelogram TCDE.

SFAT EM

H C B C D K

As in the former case, produce BD both ways. Make CB, BG, GH equal and also CD, DK and KL equal to one another. Draw HS, GF parallel to BA and LP, KM parallel to DE. Produce AE both ways to S P. It may be shown that GS, BF, DM, KP, CS, CP are parallelograms.

But the parallelograms GS, BF and BT are equal (P. 14) and so are the parallelograms DT, DM and ML.

Because there are four magnitudes, viz., the two bases BC and CD, and the two parallelograms BT and DT, and of the base BC and the parallelogram BT, the first and the third, any equimultiples whatever have been taken, viz., the base CH and the parallelogram CS; and of the base