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PROPOSITION 43. THEOREM.

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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.

Because ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal to the triangle ADC.

[I. 34. Again, because AEKH is a parallelogram, and AK its diameter, the triangle AEK is equal to the triangle BG AHK.

[I. 34. 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. [A.. 2. But the whole triangle ABC was shewn to be equal to the whole triangle ADČ. Therefore the remainder, the complement BK, is equal to the remainder, the complement KD.

[Axiom 3. Wherefore, the complements &c. Q.E.D.

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 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.

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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 ; [1. 42. produce FG to H; through A draw AH parallel to BG or EF, [I. 31. and join HB.

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. [A.c. 12. Therefore HB and FE will meet if produced ; let them meet at K. Through K draw KL parallel to EA or FH; [I. 31. and produce HA, GB to the points L, M.

Then ALKF 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.

[I. 43. But BF is equal to the triangle C.

[Construction. Therefore LB is equal to the triangle C [Axiom 1. And because the angle GBE is equal to the angle ABM,[1.15. and likewise to the angle D;

[Construction. the angle ABM is equal to the angle D. [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.

I M 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 aretogether equal to two right angles; [I.29. therefore RHG,GHM are together equalto two right angles. KH M And because at the point H in the straight line GH, the two straight lines KH, HM, on the opposite sides of it, make the adjacent angles together equal to two right angles, KH is in the same straight line with HM. [I. 14.

And because the straight line HG meets the parallels KM, FG, the alternate angles MHG, HGFare equal. [I. 29. Add to each of these equals the angle HGL; therefore the angles MHG, HGL, are equal to the angles HGF, HGL.

[Axiom 2. But MHG,HGLare together equal to two right angles; [1. 29. therefore HGF, HGL are together equal to two right angles Therefore FG is in the same straight line with GL. [I. 14.

And because KFis parallel to HG,and HG to ML,[Constr. KF is parallel to ML;

[I. 30. and KM, FL are parallels ;

[Construction. therefore KFLM is a parallelogram.

[Definition. And because the triangle ABD is equal to the parallelogram HF,

[Construction. and the triangle DBC to the parallelogram GM; [Constr. the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM.

[Axiom 2. Wherefore, the parallelogram KFLM has been described equal to the given rectilineal figure ABCD, and having the angle FKM equal to the given angle E. Q.E.F.

COROLLARY. From this it is manifest, how to a given straight line, to apply a parallelogram, which shall have an angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure; namely, by applying to the given straight line a parallelogram equal to the first triangłe ABD, and having an angle equal to the given angle ; and so on.

[I. 44.

PROPOSITION 46. PROBLEM. To describe a square on a given straight line. Let AB be the given straight line: it is required to describe a square on AB.

From the point A draw. AC at right angles to AB; [1. 11. and make AD equal to AB; [I. 3. through D draw DE parallel to AB; and through B draw BE parallel to AD.

[I. 31. ADEB shall be a square.

For ADEB is by construction a parallelogram; therefore AB is equal to DE, and AD to BE.

[I. 34. But AB is equal to AD.

[Construction. Therefore the four straight lines BA, AD, DE, EB are equal to one another, and the parallelogram ADEB is equilateral.

[Axiom 1. Likewise all its angles are right angles. For since the straight line AD meets the parallels AB, DE, the angles BAD, ADE are together equal to two right angles ;

[I. 29. but BAD is a right angle ;

[Construction. therefore also ADE is a right angle.

[A.ciom 3. But the opposite angles of parallelograms are equal. [I. 34. Therefore each of the opposite angles ABE, BED is a right angle.

[Axiom 1 Therefore the figure ADEB is rectangular; and it has been shewn to be equilateral. Therefore it is a square.

[Definition 30. And it is described on the given straight line AB. Q.E.F.

COROLLARY. From the demonstration it is manifest that every parallelogram which has one right angle has all its angles right angles.

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