24 of this.

† 44 of this.

14 of this,

PROPOSITION XLV.

PROBLEM.

To make a Parallelogram equal to a given Rightlined Figure, in a given Right-lined Angle. LET ABCD be the given Right-lin❜d Figure, and E the Right-lin'd Angle given. It is required to make a Parallelogram equal to the Rightlined Figure A B CD, in an Angle equal to E.

Let D B be joined, and make the Parallelogram FH equal to the Triangle A B D, in an Angle HK F, equal to the given Angle E.

Then to the Right Line GH apply + the Parallelogram G M, equal to the Triangle DBC in an Angle G H M, equal to the Angle E.

Now, becaufe the Angle E is equal to H KF, or GHM, the Angle HK.F fhall be equal to G HM, add K H G to both; and the Angles HKF, KHG, are, together, equal to the Angles K HG, GHM. 29 of this. But HKF, KHG, are ‡, together, equal to two Right Angles. Wherefore, likewife, the Angles KHG, GHM, fhall be equal to two Right Angles: And fo, at the given Point H in the Right Line G H, two Right Lines K H, HM, not drawn on the fame Side, make the adjacent Angles, both together, equal to two Right Angles; and confequently K H, H M*, make one ftrait Line. And because the Right Line HG falls upon the Parallels K M, F G, the alternate Angles MHG, HGF, are equal. And if HGL be added to both, the Angles M HG, HG L, together, are equal to the Angles H GF, HGL, together. 29 of this. But the Angles M HG, HG L, are together equal to two Right Angles. Wherefore, likewife, the Angles HGF, HGL, are together equal to two Right Angles; and fo, FG, GL, make one ftrait Line. And fince KF is equal and parallel to H G, as like30 of this. wife HG to ML, KF fhall be + equal and parallel to M L, and the Right Lines K M, FL, join them. 33 of this. Wherefore K M, FL, are ‡ equal and parallel. There

*

fore K FLM is a Parallelogram. But fince the Triangle A B D is equal to the Parallelogram H F, and

the

the Triangle DBC to the Parallelogram GM; therefore the whole Right-lined Figure ABCD will be equal to the whole Parallelogram K FLM. KF Therefore, the Parallelogram K F L M is made equal to the given Right-lined Figure A B C D, in an Angle FK M, equal to the given Angle E: which was to be done.

Coroll. It is manifeft from what has been faid, how to apply a Parallelogram to a given Right Line, equal to a given Right-lin'd Figure in a given Rightlined Angle.

PROPOSITION XLVI.

PROBLEM.

To defcribe a Square upon a given Right Line.

L

ETAB be the Right Line given, upon which it
is required to defcribe a Square.

*

Draw A C at Right Angles to A B from the Point * 11 of this. A given therein; make † AD equal to A B, and thro' † 3 of this. the Point D draw t DE parallel to A B; alfo thro' t 11 of tbiso B draw B E parallel to A D.

Then A DEB is a Parallelogram; and fo A B is 34 of this. equal to D E, and AD to BE. But B A is equal to AD. Therefore, the four Sides BA, AD, BE, ED, are equal to each other.

And fo the Parallelogram ADEB is equilateral: I fay, it is likewife equiangular. For becaufe the Right Line A D falls upon the Parallels A B, D E, the Angles BAD, A DE, are + equal to two Right Angles. † 29 of this. But BAD is a Right Angle. Wherefore ADE is

alfo a Right Angle; but the oppofite Sides and oppo

fite Angles of Parallelograms are equal. Therefore, † 34 of this. each of the oppofite Angles A BE, BED, are Right Angles; and confequently A D BE is a Rectangle : But it has been proved to be equilateral. Therefore, it is neceffarily a Square, and is defcribed upon the Def. 30. Right Line A B; which was to be done.

*

Coroll. Hence every Parallelogram that has one Right

Angle, is a Rectangle.

PRO

46 of this.

31 of this.

+ Def. 30.

PROPOSITION XLVII.

THEOREM.

In any Right-angled Triangle, the Square defcribed upon the Side, fubtending the Right Angle, is equal to both the Squares defcribed upon the Sides, containing the Right Angle.

LET ABC be a Right-angled Triangle, having the Right Angle BA C. I fay the Square, defcribed upon the Right Line B C, is equal to both the Squares, defcribed upon the Sides B A, A C.

*

For, delcribe upon B C the Square B DEC, and on BA, AC, the Squares G B, HC; and through the Point A draw A L parallel to || BD, or CE; and let A D, F C, be joined.

Then, because the Angles BA C, BAG, † are Right ones, two Right Lines A G, AC, at the given Point A, in the Right Line BA, being on contrary Sides thereof, make the adjacent Angles equal to two 14 of this. Right Angles. Therefore CA, AG, make

• Ax. 2.

one

ftrait Line, by the fame Reafon AB, AH, make one ftrait Line. And fince the Angle D B C is equal to, the Angle F B A, for each of them is a Kight one, add A BC, which is common, and the whole Angle DBA is equal to the whole Angle F B C. And fince the two Sides A B, BD, are equal to the two Sides F B, BC, each to each, and the Angle D BA † 4 of this, equal to the Angle FBC; the Bafe A D will be f equal to the Bafe F C, and the Triangle A B D equal to the Triangle FBC; but the Parallelogram BL is double to the Triangle A BD; for they have the fame Bale D B, and are between the fame Parallel. 141 of this. B D, AL. The Square GB is alfo double to the Triangle FBC; for they have the fame Bafe F B, and are in the fame Parallels F B, G. C. But Things that are the Doubles of equal Things,, are equal to each other. Therefore the Parallelogram B L is equal to the Square G B. After the fame manner, A E, B K, being joined, we prove that the Parallelogram CL is equal to the Square H C. Therefore the whole Square BDEC is equal to the two Squarts G B, HC. But the Square B D E C is defcribed on the

* Ax. 6.

*