Then the triangle ABC is equal to the triangle GEF, because they are on equal bases BC, EF, and between the same parallels.

[I. 38.

But the triangle ABC is equal to the triangle DEF. [Hyp. 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.

[Axiom 1.

In the same manner it can be shewn that no other straight line through A but AD is parallel to BF; therefore AD is parallel to BF. Wherefore, equal triangles &c. Q.E.D.

PROPOSITION 41. THEOREM.

If a parallelogram and a triangle be on 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 on the same base BC, and between the same parallels BC, AE: the parallelogram ABCD shall be double of the triangle EBC.

Join AC.

Then the triangle ABC is equal to the triangle EBC, because they are on the same base BC, and between the same parallels BC, AE. [I. 37.

But the parallelogram ABCD is double of the triangle ABC,

D

because the diameter AC bisects the parallelogram. [I. 34. Therefore the parallelogram ABCD is also double of the triangle EBC.

Wherefore, if a parallelogram &c. Q.E.D.

PROPOSITION 42. 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.

Bisect BC at E: [I. 10. join AE, and at the point E, in the straight line EC, make the angle CEF equal to D; [I. 23. through A draw AFG parallel to EC, and through C draw CG parallel to

EF.

[I. 31.

Therefore FECG is a parallelogram.

And, because BE is equal to EC,

[Definition. [Construction.

the triangle ABE is equal to the triangle AEC, because they are on equal bases BE, EC, and between the same parallels BC, AG.

[I. 38.

Therefore the triangle ABC is double of the triangle AEC. But the parallelogram FECG is also double of the triangle AEC, because they are on the same base EC, and between the same parallels EC, AG.

[I. 41. Therefore the parallelogram FECG is equal to the triangle ABC;

[Axiom 6.

and it has one of its angles CEF equal to the given angle [Construction.

D.

Wherefore a parallelogram FECG has been described equal to the given triangle ABC, and having one of its angles CEF equal to the given angle D. Q.E.F.

PROPOSITION 43. THEOREM.

The complements of the parallelograms which are about the diameter of any parallelogram, are equal to one another.

Let ABCD be a parallelogram, of which the diameter is AC; and EH, GF parallelograms about AC, that is, through which AC passes; and BK, KD the other parallelograms which make up the whole figure ABCD, and which are therefore called the complements: the complement BK shall be equal to the complement KD.

For the same reason the triangle KGC is equal to the triangle KFC.

Therefore, because the triangle AEK is equal to the triangle AHK, and the triangle KGC to the triangle KFC; the triangle AEK together with the triangle KGC is equal to the triangle AHK together with the triangle KFC. [Ax. 2. But the whole triangle ABC was shewn to be equal to the whole triangle ADC.

Therefore the remainder, the complement BK, is equal to the remainder, the complement KD.

Wherefore, the complements &c. Q.E.D.

[Axiom 3.

PROPOSITION 44. 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.

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.

Make the parallelogram BEFG equal to the triangle C, and having the angle EBG equal to the angle D, so that BE may be in the same straight line with AB ; [I. 42. produce FG to H;

through A draw AH parallel to BG or EF,

and join HB.

[I. 31.

Then, because the straight line HF falls on the parallels AH, EF, the angles AHF, HFE are together equal to two right angles.

[I. 29. Therefore the angles BHF, HFE are together less than two right angles.

But straight lines which with another straight line make the interior angles on the same side together less than two right angles will meet on that side, if produced far enough. [4x.12. Therefore HB and FE will meet if produced;

let them meet at K.

Through K draw KL parallel to EA or FH; and produce HA, GB to the points L, M.

[I. 31.

Then HLKF is a parallelogram, of which the diameter

is HK; and AG, ME are parallelograms about HK; and

LB, BF are the complements.

Therefore LB is equal to BF.

But BF is equal to the triangle C.

[I. 43.

[Construction.

Therefore LB is equal to the triangle C

[Axiom 1.

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.

[Construction. [Axiom 1.

Wherefore to the given straight line AB the parallelogram LB is applied, equal to the triangle C, and having the angle ABM equal to the angle D. Q.E.F.

PROPOSITION 45. 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.

Join DB, and describe the parallelogram FH equal to the triangle ADB, and having the angle FKH equal to the angle E;

[I. 42. and to the straight line GH apply the parallelogram GM equal to the triangle DBC, and having the angle GHM equal to the angle E. [I. 44. The figure FKML shall be the parallelogram required. Because the angle E is equal to each of the angles FKH, GHM, [Construction. the angle FKH is equal to the angle GHM. [Axiom 1. Add to each of these equals the angle KHG; therefore the angles FKH, KHG are equal to the angles KHG, GHM. [Axiom 2. But FKH, KHG are together equal to two right angles; [1.29. therefore KHG,GHM are together equal to two right angles.