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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.

* 34. 1. Again, because EKHA is a parallelogram, the diameter of which is AK, the triangle AEK is equal * to the tri- * 34. 1. angle AHK: for a like reason, the triangle KGC is equal to the triangle KFC. But the triangle AEK was proved equal to the triangle AHK, and the triangle KGC to KFC; therefore the triangle AEK together with the triangle KGC is equal to the triangle AHK together with the triangle KFC: and the whole triangle ABC was proved equal to the whole ADC; therefore the remaining complement BK is equal * to * 3 Ax. the remaining complement KD. Wherefore the complements, &c. Q. E. D.

PROP. XLIV. PROB.
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 parallel

* 42. 1. ogram 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, that is, in AB produced, and produce FG to H; and through A draw* AH * 31. 1. parallel to BG or EF, and join HB. Then, because the straight line Hr falls upon the parallels AH, EF,

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* 29 1. the angles AHF, HFE, are together equal* to two

right angles; therefore the angles BHF, HFE are less than two right angles: but straight lines, which, with

another straight line, make the interior angles upon * 12 Ax. the same side less than two right angles, will meet* if

produced far enough: therefore HB, FE will 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 HLKP is a parallelogram, of which the diameter is HK, and AG, ME are the parallelograms about HK; and LB, BF are the complements; therefore LB is equal * to BF: but BF is equal

to the triangle C; wherefore LB is equal to the tri* 15. 1. angle C; and because the angle GBE is equal * to the

angle ABM, and likewise to the angle D; the angle ABM is equal to the angle D. Therefore, to the straight line AB, the parallelogram LB is applied equal to the triangle C, having the angle ABM equal to the angle D. Which was to be done.

* 43. 1.

PROP. XLV. PROB. 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 that shall be equal to ABCD, and have

an angle equal to E. * 42. 1.

Join DB; and describe* the parallelogram FH equal to the triangle ADB, and having the angle HKF

equal to the angle E; and to the straight line GH * 44. 1. apply* the parallelogram GM equal to the triangle

DBC, having the angle GHM equal to the angle E: then FKLM shall be the parallelogram required.

Because the angle E is equal to each of the angles FKH, GHM, the angle FKH is equal to GHM; add

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to each of these equals the angle KHG; therefore the angles FKH, KHG are equal to the angles KHG, GHM; but FKH, KHG are equal*

* 29. 1. to two right angles: therefore KHG, GHM are also equal to two right angles; and because at the point H, in the straight line GH, the two straight lines KH, HM, upon the opposite sides of it, make the adjacent angles equal to two right angles, KH is in the same straight line* with HM; and because the straight line * 14. 1. HG meets the parallels KM, FG, the alternate angles MHG, HGF are equal.:* add to each of these equals * 29. 1. the angle HGL; therefore the angles MHG, HGL, are equal to the angles HGF, HGL: but the angles MHG, HGL are equal * to two right angles; where- * 29. 1. fore the angles HGF, HGL are also equal to two right angles, and FG is therefore in the same straight line * * 14. 1. with GL: and because KF is parallel to HG, and HG to ML, KF is parallel * to ML: and KM, FL are * 30. 1. parallels; wherefore KFLM is a parallelogram; and because the triangle ABD is equal to the parallelogram HF, and the triangle DBC to the parallelogram GM; the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM; therefore the parallelogram KFLM has been described equal to the given rectilineal figure ABCD, having the angle FKM equal to the given angle E. Which was to be done.

Cor. 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, viz. by applying * * 44. 1. to the given straight line a parallelogram equal to the first triangle ABD, and having an angle equal to the given angle.

* 1l. 1.

* 3. 1.

PROP. XLVI. PROB.
To describe a square upon a given straight line.
Let AB be the given straight line; it is required to
describe a square upon AB.

From the point A draw* AC at right angles to AB;

and make* AD equal to AB, and through the point D * 31. 1. draw De parallel * to AB, and through B draw BE

parallel to AD; then ABCD shall be the square re

quired. Since ADEB is a parallelogram, therefore * 34. 1. AB is equal * to DE, and AD to BE:

but BA is equal to AD; therefore the
four straight lines BA, AD, DE, EBD
are equal to one another, and the pa-
rallelogram ADEB is equilateral.
Likewise all its angles are right an-
gles; for, since the straight line AD A
meets the parallels AB, DE, the angles BAD, ADE
are equal * to two right angles; but BAD is a right

angle; therefore also ADE is a right angle; but the * 34. 1. opposite angles of parallelograms are equal;* there

fore each of the opposite angles ABE, BED is a right angle; therefore the figure ADEB is rectangular, and it has been demonstrated that it is equilateral; it is therefore a square, and it is described upon the given straight line AB. Which was to be done.

Cor. Hence every parallelogram that has one right angle has all its angles right angles.

* 29. 1.

PROP. XLVII. THEOR.
In any right-angled triangle, the square which is de-

scribed upon the side subtending the right angle, is
equal to the squares described upon the sides which
contain the right angle.

Let ABC be a right-angled triangle having the right angle BAC; the square described upon the side

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BC shall be equal to the squares described upon
BA, AC.

On BC describe* the square BDEC, and on BA, * 46. 1. AC the squares GB, HC; and through A draw* AL * 31. 1. parallel to BD, or CE, and join AD, FC; then, because each of the angles BAC, BAG is a right angle *, * 30 Def. the two straight lines AC, AG upon the opposite sides of AB, make with it at the point A the adjacent angles equal to two right angles; therefore CA is in the same

* 14. 1. straight line* with AG; for the same reason, AB and AH are in the same straight line; and because the angle DBC is equal to the angle FBA, each of them being a right angle, add to each of these equals the angle ABC, therefore the whole angle DBA is equal* to the whole * 2 Ax. FBC; and because the two sides AB, BD are equal to the two FB, BC, each to each, and the angle ABD equal to the angle FBC; therefore the triangle ABD is equal* to the triangle FBC: but the parallelogram * 4. 1. BL is double* of the triangle ABD, because they are

* 41. 1. upon the same base BD, and between the same parallels BD, AL; and the square GB is double of the triangle FBC, because these also are upon the same base FB, and between the same parallels FB, GC. But the doubles of equals are equal* to one another: therefore * 6 Ax. the parallelogram BL is equal to the square GB: and, in like manner, by joining AE, BK; it can be demonstrated that the parallelogram CL is equal to the square HC: therefore the whole square BDEC is equal to the two squares GB, HC; and the square BDEC is described upon the straight line BC, and the squares GB, HC upon BA, AC: therefore the square

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