If a straight line be divided into any two parts, the squares of the whole line, and of one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square of the other part.

A

C B

Let the straight line AB be divided into any two parts in the point C; the squares of AB, BC, are equal to twice the rectangle AB.BC, together with the square of AC, or AB+BC-AB.BC+AC": Upon AB describe (46. 1.) the square ADEB, and construct the figure as in the preceding propositions: Because AG-GE, AG÷CK GE+CK, that is AKCE, and therefore, AK+CE=2AK. But AK+CEgnomon AKF+CK; and therefore, AKF +CK=2AK2AB.BK-2AB.BC, because BK=(Cor. 4. 2.) BC. Since then, H AKF+CK=2AB.BC, AKF+CK+HF =2AB.BC+HF; and because AKF+ HF-AE=AB3, AB3+CK=2AB.BC+ HF, that is, (since CK-CB, and HF AC,) AB+ CB2AB.BC+AC. Wherefore, if a straight line, &c. Q. E.D.

Otherwise,

D

FE

"Because ABAC+BC+2AC.BC (4. 2.), adding BC to both, AB+BC= AC'+2BC+2AC.BC:

But BC2+AC.BC AB.BC (3. 2.); and

11

therefore, 2BC2+2AC.BC÷2AB.BC; and A

therefore AB+BC AC2+2AB.BC."

C

B

"COR. Hence the sum of the squares of any two lines is equal to twice the rectangle contained by the lines together with the square of the difference of the lines."

PROP. VIII. THEOR.

If a straight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the square of the other part, is

H

equal to the square of the straight line which is made up. of the whole and the first-mentioned part,

Let the straight line AB be divided into any two parts in the point C; four times the rectangle AB.BC, together with the square of AC, is equal to the square of the straight line made up of AB and BC together.

A

CBD

GK

N

P

R

0

Produce AB to D, so that BD be equal to CB, and upon AD describe the square AEFD; and construct two figures such as in the preceding. Because GK is equal (34. 1.) to CB, and CB to BD, and BD to KN, GK is equal to KN. For the same reason, PR is equal to RO; and because CB is equal to BD, and GK to KN, the rectangles CK and BN are equal, as also the rectangles GR and RN: But CK is equal (43. 1.) to RN, because they are the complements of the parallelogram CO; therefore also BN is equal to GR; and the four rectangles BN, CK, GR, RN are therefore equal to one another, and so CK+BN+GR+RN=4CK. Again, because CB is equal to BD, and BD equal (Cor. 4. 2.) to BK, that is to CG; and CB equal to GK, that (Cor. 21. 2.) is, to GP; therefore CG is equal to GP; and because CG is M equal to GP, and PR to RO, the rectangle AG is equal to MP, and PL to X 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 AG +MP+PL+RF-4AG. And it was demonstrated, that CK+BN+GR+RN=4CK; wherefore, adding equals to equals, the whole gnomon AOH=4AK. Now AK-AB.BK. AB.BC, and 4AK=4AB.BC; therefore, gnomon AOH=4AB.BC; and adding XH, or (Cor. 4. 2.) AC, to both, gnomon AOH+XH= 4AB.BC+AC. But AOH+XH-AF-AD; therefore AD4AB.BC+AC2. Now AD is the line that is made up of AB and BC, added together into one line: Wherefore, if a straight line, &c. Q. E. D.

1

E

H L F

"COR. 1. Hence, because AD is the sum, and AC the difference of the lines AB and BC, four times the rectangle contained by any two lines together with the square of their difference, is equal to the square of the sum of the lines.”

"Cor. 2. From the demonstration it is manifest, that since the square of CD is quadruple of the square of CB, the square of any line is quadruple of the square of half that line."

Otherwise :

"Because AD is divided any how in C (4.2.), AD2=AC®+CD' +2CD.AC. But CD=2CB: and therefore CD=CB3+BD2+ 2CB.BD (4.2.)=4CB', and also 2CD.AC-4CB.AC; therefore,

ADAC+4BC2+4BC.AC. Now BC+BC.AC-AB.BC (3. 2.) and therefore AD-AC+4AB.BC. Q, E, D,"

PROP. IX. THEOR.

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 anequal parts: The squares of AD, DB are together double of the squares of AC, CD.

E

F

C D

B

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 angle. For the same reason each of the angles CEB, EBC is half a right angle: and therefore the whole AEB is a right angle: And because the angle GEF is half a right angle, and EGF a right angle, for it is equal (29. 1.) to the interior and opposite angle ECB, the remaining angle EFG is half a right A 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 right angle; and FDB a right angle, for it is equal (29. 1.) to the interior 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. Now, because AC=CE,AC®-CE®, and AC+CE2AC". But (47. 1.) AE-AC+CE; therefore AE-2AC. Again, because EG-GF, EG GF, and EG2+GF =2GF. But EF EGGF; therefore, EF-2GF-2CD2, because (34. 1.) CD=GF. And it was shown that AE-2AC2; therefore AE+EF-2AC+2CD. But (47. 1.) AFAE+ EF, and AD3+DF2=AF2, or AD+DB-AF; therefore, also AD+DB=2AC+2CD. Therefore, if a straight line, &c. Q. E. D.

Otherwise:

3

"Because AD (4. 2.) AC+CD+2AC.CD, and DB2+2BC, CD (7. 2.) BC+CD AC+CD, by adding equals to equals, AD+BD + 2BC.CD2AC2 + 2CD3 + 2AC.CD, and therefore taking away the equal rectangles 2BC.CD and 2AC.CD, there re mains AD+DB=2AC+2CD","

PROP. X. THEOR.

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, CĎ.

From the point C draw (11.1.) CE at right angles to AB, and make it equal to AC or CB; 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 therefore 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, (Cor. 29, 1.) if produced far enough: Therefore EB, FD will meet, if produced towards, B, D; 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 because 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, because 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 the angle DBG; wherefore also the side DB is equal (6. 1.) to the side DG. Again, because EGF is half a right angle, and the angle at Fa right angle, being 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

E

F

the side GF is equal (6. 1.) to

the side FE. And because EC

B

= CA, EC2 + CA3 = 2CA3. A

C

D

Now AE (47. 1.) AC+CE;

therefore, AE-2ÁC3. Again,

G

because EF FG, EF-FG", and EF+FG-2EF. But EG (47. 1.) EF+FG; therefore EG-2EF; and since EF-CD, EG 2CD. And it was demonstrated, that AE-2AC; therefore, AE+EG=2AC+2CD". Now, AG2=AE+EG, wherefore AG 2AC+2CD. But AG (47, 1.)-AD+DG=AD+DB", because DG-DB: Therefore, AD+DB=2AC+2CD". Wherefore, if a straight line, &e. Q. E. D.

PROP. XI. PROB.

To divide a given straight line into two parts, so that the rectangle contained by the whole, and one of the parts, may 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 (46. 1.) the square FGHA; AB is divided in H, so that the rectangle AB, BH is equal to the square of AH.

G

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, be- F cause 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 A is equal to the square of AB. Now the figure FK is the rectangle CF.FA, for AF is equal to FG; and AD is the square of AB; therefore FK is equal to AD: take E away the common part AK, and the remainder FH is equal to the remainder HD. But HD is the rectangle 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: C

K

H

B

D

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. PROP. XII. THEOR.

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 between the perpendicular and the obtuse angle.

Let ABC be an obtuse angled triangle, having the obtuse angle