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+CK=2AK= 2AB.BK = - 2AB.BC, because BK = (Cor. 4. 2.) BC. Since then, AKF+CK=2AB.BC, AKF+CK+HF=2AB.BC+HF; and because AKF+HF=AE=AB2, AB2+CK=2AB.BC+HF, that is, (since CK=CB?, and HF=AC?,) AB2+ CB2=2AB.BC+AC2.

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."

Cor.

SCHOLIUM.

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In this proposition, let AB be denoted by a, and the segments AC and CB by b and c;

then a=62+2bc+c2; adding c to each member of this equality, we shall have,

a2+c=62+2bc+2c; ..a2+ca=b2+2c(b+c),

or a-+c=2ac+62. Cor. From this proposition it is evident, that the square described on the difference of two lines is equivalent to the sum of the squares described on the lines respectively, minus twice the rectangle contained by the lines. For ac=b; therefore, by involution, aa-2ac+c=62. This may be also derived from the above algebraical equality, by transposition.

PROP. VIII. THEOR.

If a straight line be divided into any two parts, four times the rectangle con

tained by the whole line, and one of the parts, together with the square of the other part, is 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.

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 (28. 1.) to CB, and CB to BD, and BD to KN, GK is equal to KN. For the same reason, PR

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A is equal to RO; and because CB is equal

D to BD, and GK to KN, the rectangles CK and BN are equal, as also the rectangles

G KO

M GR and RN: But CK is equal (36. 1.)

N to RN, because they are the complements

PV R
X

lo 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

E

Η LG F

(Cor. 4. 2.) to BK, that is, to CG; and CB equal to GK, that 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 (36. 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+AC2. But AOH+XH=AF = AD2; therefore AD2=4AB.BC+AC2.

“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 druple of the square of half that line.”

SCHOLIUM. In this proposition, let the line AB be denoted by a, and the parts AC and CB by c and b; then AD=c+2b. Now, since a=b+c, multiplying both members by 4b, we shall have

4ab=462 +4bc; and adding c2 to each member of this equality, we shall have,

4ab+c=c2 +4bc+462, or 4ab+c=(c+20).

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66

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 unequal parts ; The squares of AD, DB are together double of the squares AC, CD.

From the point C draw (Prob. 6. 1.) CE at right angles to AB, and make it equal to AC or CB, and join EA, EB; through D draw (Prob. 13. 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 (3. 1.) to the

E angle AEC; and because the angle ACE is a right angle, the two others AEC,

G EAC together make one right angle (Cor. 4. 25. 1.); and they are equal to one another; each of them therefore is half of a right angle. For the same reason each A

C D B

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 (21. 1.) to the interior and opposite angle ECB, the remaining angle EFG is half a right angle; therefore the angle GEF is equal to the angle EFG, and the side EG equal (4. 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 (21. 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 (4. 1.) the side DB. Now, because AC=CE, AC=CE?, and ACP+CE=2AC2. · But (37. 1.) AE?= AC2+CE2; therefore AE?=2AC2. Again, because EG=GF, EGP=GF, and EGP+GF2=2GF. But EF=ĚG2+GF2; therefore, EF=2GF2 =2CD2, because (28. 1.) CD=GF. And it was shown that AE2=2AC2; therefore AE2+ ÈF2=2AC2+2CD2. But (37. 1.) AFP=AE? + EF?, and AD2+DF=AF?, or AD2 + DB2=AF2; therefore, also, AD2+DB2= 2AC2+2CD2

SCHOLIUM.
This property is evident from the algebraical expression,

(a+b)2+(ab)2=2a2 +262; where a denotes AC, and b denotes CD; hence, atb=AD, ab=DB.

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, CD.

From the point C draw (Prob. 6. 1.) CE at right angles to AB, and make it equal to AC or CB; join AE, EB; through E draw (Prob. 13. 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 (21. 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 (22. 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 (3. 1.) to

F the angle EAC; and the angle ACE is a right angle ; therefore each of the angles CEA, EAC is half a right angle (Cor. 4. 25. 1.); For the same reason, each of the AS

B 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 (8. 1.) half a right angle, for they are vertically opposite : but BDG is a right angle, because it is equal (21. 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 (4. 1.) to the side DG. Again, because EGF is half a right angle, and the angle at F a right angle, being equal (28. 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 (4. 1.) to the side FE And because EC=CA, EC2+CAP=2CA?. Now A E=(37. 1.) AC2+ CE?; therefore, AE2=2AC2. Again, because EF=FG, EF2=FG2, and EF2 +FGP=2EF2. But EG?= (37. 1.) EF2+FG2; therefore EG?= 2EF2; and since EF=CD, EGP=2CD2. And it was demonstrated, that AE2=2AC2; therefore, AE + EGP=2AC2+2CD2 Now, AGP=AE2+ EG?, wherefore AGP=2AC2+2CD2 But AGP (37. 1.) =AD2+DG2= AD? + DB, because DG=DB: Therefore, AD)2+DB2=2AC2+2CD2.

SCHOLIUM.

Let AC be denoted by a, and BD, the part produced, by b; then AD= 2a+b, and CD=a+b.

Now, (2a+3)2 + b2 = 4a2+4ab+262 ; but 4a2+4ab+2b2=2a2+2 (at b)2; hence, (2a+)2 +12=2a2+2(a+b), and the proposition is evident from this algebraical equality.

PROP. XI. 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 intcrcepted between the perpendicular and the ottuse angle.

Let ABC be an obtuse angled triangle, having the obtuse angle ACB, and from the point A let AD be drawn (Prob. 7. 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, BD2 = (4. 2.) BC2+CD2+2BC.CD; add AD2 to both: Then BD2+ AD2

BC2 + CD2 + AD2 + 2BC.CD.
But ABP = BD2+ AD2, (37. 1.); and
AC=CD2+AD? (37. 1.); therefore,-
AB2 BC2 + AC2 + 2BC.CD; that
is, AB2 is greater than BC2+AC2 by B

D 2BC.CD.

C

PROP. XII. THEOR.

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 perpendicular (Prob. 7. 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

A ABC ; and because the straight line CB is divided into two parts in the point D (7. 2.), BC2 + BD2 = 2BC.BD+CD2. Add to each AD2; then BC2+BD2+ AD2=2BC.BD+CD2 +AD2. But BD2 +AD2 = AB?, and CD2 + DAP=AC2 (37. 1.); therefore BC2+AB2=2BC. BD+AC2; that is, ACP is less than BC2+AB2 by 2BC.BD.

B D Secondly, Let AD fall without the triangle ABC ;* Then because the angle at D is a right angle, the angle ACB is greater (9. 1.) than a right angle, and AB2=(11. 2.) AC2+BC2+2BC.CD. Add BC2 to each ; then AB?+ BC2-AC2+2BC2+2BC.CD. But because BD is divided into two parts in C, BC2+BC.CD=(3. 2.) BC.BD, and 2BC2+2BC.CD= 2BC.BD: therefore AB2+ BC=AC2+2BC.BD, and AC2 is less than AB2+ BC?, by 2BD.BC.

Lastly, Let the side AC be perpendicular to BC; then is BC the straight line between a los resto oth

organise se A the perpendicular and the acute angle at B; and it is manifest that (37. 1.) AB2+BC2= AC2+2BC=AC2+2BC.BC.

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PROP. A. THEOR.

If one side of a triangle be bisected, the sum of the squares of the other two sides

is double of the square of half the side bisected, and of the square of the line drawn from the point of bisection to the opposite angle of the triangle.

Let ABC be a triangle, of which the side BC is bisected in D, and DA drawn to the opposite angle; the squares of BA and AC are together double of the squares of BD and DA.

* See figure of the last Proposition.

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