M

X

CBD

G

E HLF

equal (Cor. 4.2.) to BK, that is, to CG, and CB; equal to GK, that (Cor. 4. 2.) is, to GP; therefore CG is equal to GP: And because CG is equal to GP, and PR to RO, the rectangle AG is equal to MP, and PL to RF: But MP is equal (43.1.) to PL, because they are the complements of the parallelogram ML; wherefore AG is equal also to RF: Therefore, the four rectangles AG, MP, PL, RF, are equal to one another, and so are quadruple of one of them AG. And it was demonstrated that the four CK, BN, GR, and RN are quadruple of CK. Therefore the eight rectangles which contain the gnomon AOH, are quadruple of AK; and because AK is the rectangle contained by AB, BC, for BK is equal to BC, four times the rectangle AB, BC, is quadruple of AK: But the gnomon AOH was demonstrated to be quadruple of AK: therefore four times the rectangle AB, BC, is equal to the gnomon AOH. To each of these add XH, which is equal (Cor. 4 2) to the square of AC; Therefore four times the rectangle AB, BC, togeer with the square of AC, is equal to the gnomon AOH and the square XH: But the gnomon AOH and XH make up the figure AEFD, which is the square of AD: Therefore four times the rectan gle AB, BC, together with the square of AC, is equal to the square of AD, that is, of AB and BC added together in one straight line. Wherefore, if a straight line, &c. Q. E. Ď.

Proposition IX. Theorem.

If a straight line be divided into two equal, and also into two unequal parts; the squares of the two unequal parts are together double of the square of half the line, and of the square of the line between the points of section.

Let the straight line AB be divided at the point C into two equal, and at D into two unequal parts: The squares of AD, DB, are together double of the squares of AC, CD.

From the point C draw (11. 1.) CE at right angles to AB, and make it equal to AC, or CB, and join EA, EB; through D draw (31. 1.) DF parallel to CE, and through F draw FG parallel to AB; and join AF. Then because AC is equal to CE, the angle EAC is equal (5. 1.) to the angle AEC; and because the angle ACE, is a right angle, the two others AEC, EAC together make one right angle; (32. 1.) and they are equal to one another; each of them therefore is half of a right angis. For the same reason each of the angles CEB, 4BC is half a right angle; and therefore the whole AEB, is a right angle: And because he angle GEF, is half a right angle, and EGF a right angle, for it is equal (29. 1.) to .he interior and opposite angle ECB, the renaining angle EFG is half a right angle; therefore the angle GEF is equal to the angle EFG, and the side EG equal (6.1.) to the side GF: Again, because the angle at B is half a

A

right angle, and FDB a right angle, for it is equal (29. 1.) to the inte rior and opposite angle ECB, the remaining angle BFD, is half a right angle; therefore the angle at B is equal to the angle BFD, and the side DF to (6.1.) the side DB: And because AC is equal to CE, the square of AC is equal to the square of CE; therefore the squares of AC, CE, are double of the square of AC: But the square of EA is equal (47. 1.) to the squares of AC, CE, because ACE is a right angle; therefore the square of EA is double of the square of AC: Again, because EG is equal to GF, the square of EG is equal to the square of GF; therefore the squares of EG, GF are double of the square of GF; but the square of EF is equal to the squares of EG, GF; therefore the square of EF is double of the square GF; and GF is equal (34. 1.) to CD; therefore the square of EF is double of the square of CD: But the square of AE is likewise double of the square of AC; therefore the squares of AE, EF are double of the squares of AC, CD: And the square of AF is equal (47. 1.) to the squares of AE, EF, because AEF is a right angle; therefore the square of AF is double of the squares of AC, CD: But the squares of AD, DF, are equal to the square of AF, because the angle ADF is a right angle; therefore the squares of AD, DF are double of the squares of AC, CD: And DF is equal to DB; therefore the squares of AD, DB are double of the squares of AC, CD. If therefore a straight line, &c. Q. E. D.

Proposition X. Theorem.

If a straight line be bisected, and produced to any point, the square of the whole line thus produced, and the square of the part of it produced, are together double of the square of half the line bisected, and of the square of the line made up of the half and the part produced. Let the straight line AB be bisected in C and produced to the point D; the squares of AD, DB are double of the squares of AC, CD. From the point C draw (11. 1.) CE at right angles to AB: And make it equal to AC, or CB, and join AE, EB; through E draw (31. 1.) EF parallel to AB, and through D draw DF parallel to CE: And because the straight line EF meets the parallels EC, FD, the angles CEF, EFD are equal (29. 1.) to two right angles; and there. fore the angles BEF, EFD are less than two right angles; but straight lines which with another straight line make the interior angles upon the same side less than two right angles, do meet (12. Ar, if produced far enough: Therefore EB, FD shall meet if produced towards BD: Let them meet in G, and join AG: Then, because AC is equal to CE, the angle CEA is equal (5. 1.) to the angle EAC; and the angle ACE is a right angle; therefore each of the angles CEA, EAC is half a right angle. (32.1.) For the same reason, each of the angles CEB, EBC is half a right angle; therefore AEB is a right angle: And be cause EBC is half a right angle, DBG is also (15. 1.) half a right angle, for they are vertically opposite; but BDG is a right angle be cause it is equal (29. 1.) to the alternate angle DCE; therefore the remaining angle DGB is half a right angle, and is therefore equal to

E

B

the angle DBG; wherefore also the side BD is equal (6. 1.) to the side DG. Again, because EGF is half a right angle, and that the angle at F is a right angle, because it is equal (34. 1.) to the opposite angle ECD, the remaining angle FEG is half a right angle, and equal to the angle EGF; wherefore also the side GF, is equal (6. 1.) to the side FE. And because EC is equal to CA, the square of EC is equal tc the square of CA; therefore the squares of EC, CA are double of the square of CA: But the square of EA is equal (47. 1.) to the squares of EC, CA; therefore the square of EA, is double of the square of AC: Again, because GF is equal to FE, the square of GF is equal to the square of FE; and therefore the squares of GF, FE are double of the square of EF: But the square of EG is equal (47. 1.) to the squares of GF, FE; therefore the square of EG is double of the square of EF: And EF is equal to CD; wherefore the square of EG is double of the square of CD. But it was demonstrated, that the square of EA is double of the square of AC: therefore the squares of AE, EG, are double of the squares of AC, CD: And the square of AG is equal (47. 1.) to the squares of AE, EG; therefore the square of AG is double of the squares of AC, CD: But the squares of AD, GD are equal (47. 1.) to the square of AG; therefore the square of AD, DG, are double of the squares of AC, CD: But DG is equal to DB; therefore the squares of AD, DB, are double of the squares of AC, CD. Wherefore, if a straight line, &c. Q. E. D.

Proposition XI. Problem.

To divide a given straight line into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part.

Let AB be the given straight line; it is required to divide it into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part.

Upon AB describe (46. 1.) the square ABDC; bisect (10.1.) AC in E, and join BE; produce CA to F, and make (3. 1.) EF equal to EB, and upon AF, describe (3. 1.) the square FGHA; AB is divided in H, so that the rectangle AB, BH, is equal to the square of AH.

Produce GH to K; because the straight line AC is bisected in E, and produced to the point F, the rectangle CF, FA, together with the square of AE, is equal (6. 2.) to the square of EF: But EF is equal to EB; therefore the rectangle CF, FA, together with the square of AE, is equal to the square of EB : And the squares of BA, AE, are equal (47.1.) to the square of EB, because the angle EAB, is a right angle; therefore the rectangle CF, FA, together with the square of AE, is equal to the squares of BA, AE: Take away the square of AE, which is common to both, therefore the remaining rectangle CF, FA, is equal to the square of AB; and the figure FKs the rectangle con

E

lained by CF, FA, for AF is equal to FG; and AD is the square of AB; therefore FK is equal to AD: Take away the common part AK, and the remainder FH is equal to the remainder HD: And HD is the rectangle contained by AB, BH, for AB is equal to BD; and FH is the square of AH.

Therefore the rectangle AB, BH is equal to the square of AH: Wherefore the straight line AB is divided in H, so that the rectangle AB, BH, is equal to the square of AH. Which was to be done.

In obtuse angled triangles, if a perpendicular be drawn from any of the acute angles to the opposite side produced, the square of the side subtending the obtuse angle is greater than the squares of the sides containing the obtuse angle, by twice the rectangle contained by the side upon which, when produced, the perpendicular falls, and the straight line intercepted without the triangle between the perpendicular and the obtuse angle.

Let ABC be an obtuse-angled triangle, having the obtuse angle ACB, and from the point A let AD be drawn (12. 1.) perpendicular to BC produced: The square of AB is greater than the squares of AC, CB, by twice the rectangle BC, CD.

Because the straight line BD is divided into two parts in the point C the square of BD is equal (4. 2.) to the squares of BC, CD, and twice the rectangle, BC, CD: To each of these equals add the square of DA; and the squares of BD, DA, are equal to the squares of BC, CD, DA, and twice the rectangle BC, CD: But the square of BA is equal (47, 1.) to the squares of BD, DA, because the angle at D, is a right angle; and the square of CA is equal (47. 1.) to the squares of CD, DA:

B

Therefore the square of BA is equal to the squares of BC, CA, and twice the rectangle BC, CD; that is, the square of BA is greater than the squares of BC, CA, by twice the rectangle BC, CD. Therefore, in obtuse angled triangles, &c. Q. E. D.

Proposition XIII. Theorem.

In every triangle, the square of the side subtending any of the acute angles, is less than the squares of the sides containing that angle by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall upon it from the opposite angle, and the acute angle.

Let ABC be any triangle, and the angle at B one of its acute angles, and upon BC, one of the sides containing it, let fall the perpendiculat (12. 1.) AD, from the opposite angle: The square of AC, opposite to the angle B, is less than the squares of CB, BA, by twice the rectangle CB, BD.

First, Let AD fall within the triangle ABC; and because the

straight line CB is divided into two parts in the point D, the squares of CB, BD are equal (7.2.) to twice the rectangle contained by CB, BD, and the square of DC :

To each of these equals add the square of AD; therefore the squares of CB, BD, DA, B are equal to twice the rectangle CB, BD, and

the squares of AD, DC: But the square of AB is equal (47. 1.) to the squares of BD, DA, because the angle BDA is a right angle; and the square of AC is equal to the squares of AD, DC:

Therefore the squares of CB, BA are equal to the square of AC, and twice the rectangle CB, BD, that is, the square of AC alone is less than the squares of CB, BA by twice the rectangle CB, BD. Secondly, Let AD fall without the triangle ABC: Then, because the angle at D, is a right angle, the angle ACB is greater (16.1.) than a right angle; and therefore the square of AB is equal (12. 2.) to the squares of AC, CB, and twice the rectangle BC, CD:

B

C

To these equals add the square of BC, and the squares of AB, BC are equal to the square of AC, and twice the square of BC, and twice the rectangle BC, CD:

But because BD is divided into two parts in C, the rectangle DB, BC is equal (3.2.) to the rectangle BC, CD and the square of BC: And the double of these are equal: Therefore the squares of AB, BC, are equal to the square of AC, and twice the rectangle DB, BC: Therefore the square of AC alone is less than the squares of AB, BC, by twice the square of BC, and twice the rectangle DB, BC. Lastly let the side AC be perpendicular to BC; then is BC the straight line between the perpendicular and the acute angle at B; and it is manifest, that the squares of AB, BC, are equal (47. 1.) to the square of AC and twice the square of BC: Therefore, in every triangle, &c. Q. E. D.

Proposition XIV. Problem.

To describe a square that shall be equal to a given rectilineal figure.

Let A be the given rectilineal figure; it is required to describe a square that shall be equal to A.

H

Describe (45. 1.) the rectangular parallelogram BCDE equal to the rectilineal figure A. If then the sides of it, BE, ED, are equal to one another, it is a square, and what was required is now done: But if they are not equal, produce one of them BE to F, and make EF equal to ED, and bisect BF in G; and from the centre G, at the distance GB, or GF, describe the semicircle BHF, and produce DE to H, and join GH:

B

E