́angle (i. Prop. 29); `but a part of it, namely, ▲ CEB, is half of a right angle; and therefore the remainder, that is, ▲ GEF, must be also half of a right angle; and consequently, since ZF is a right angle (i. Prop. 34), ▲ EGF must be also half of a right angle (i. Prop. 32), and therefore EF=FG (i. Prop. 6). And in like manner, because ▲ ADG is a right angle (i. Prop. 29), and ▲ BGD (or ▲ EGF) is half of a right angle, ▲ GBD must be also half of a right angle, and BD=DG. Now, since ▲ AEB or ▲ AEG is a right angle, as it has been demonstrated, AG2=AE2+ EG2 (i. Prop. 47); but / ACE is also a right angle, and therefore AE2= AC2+ CE2, or 2 AC2, since AC=CE. And in like manner, because F is a right angle, EG2=EF2 + FG2, or 2 EF2, since EF=FG, or 2 CD2, since EF=CD (i. Prop. 34). Therefore, AG2= 2 AC2+2 CD3. But also, since ▲ ADG is a right angle_AG2= AD2+DG2, or AD2+BD2, since BD=DG. Consequently, AD2+BD2, or the squares of the whole produced line and of the produced part, are together equal to 2 AC2+2 CD2, or double of the squares of the half line, and of the line made up of the half and the produced part.

PROP. XI. PROB.

To divide a given straight line (AB) in such a manner that the rectangle contained by the whole line and one of the parts may be equal to the square of the other part.

с

K D

Upon AB construct the square ACDB; bisect AC in E; join EB, and produce CA towards F till EF is equal to EB; upon AF construct the square AFHG; and the given line AB will be divided in G in the required manner.

E

F

H

GB

For, produce HG to K; and then, since CA A is bisected in E, and produced to F, the rectangle CF FA+EA2=EF2 (ii. Prop. 6); but EF2=EB2, since EF EB (Const.); and EB2=EA2+AB2 (i. Prop. 47), and therefore CF·FA+EA2= EA2+ AB2; and EA2 being taken from both, there remains CF·FA = AB2 (Ax. 3). But because_AF = FH, the rectangle CF•FA=CF·FH, or the rectangle CH, which is therefore equal to AB2, or the square ACDB; and if from both be taken the common part CG,

=

there will remain AFHG or AG' GD; that is to say, AG2= BD·GB, or AB·GB, since AB=BD. Therefore, the given line AB is divided in G in such a manner that the rectangle AB GB, contained by the whole line and one of the parts, is equal to AG2, or the square of the other part.

PROP. XII. THEOR.

In an obtused angled triangle (ABC), the square of the side (AC) subtending the obtuse angle, exceeds the sum of the squares of the sides (AB, BC) containing the obtuse angle, by twice the rectangle contained by either (AB) of those sides and the produced part of it (BD) intercepted between the perpendicular (CD) let fall on it from the opposite angle, and the vertex of the obtuse angle.

Since AD is divided in B, AD2 = AB2 + BD2 + 2 AB·BD (ii. Prop. 4); add to both DC2, and then AD2+DC2 =AB2+BD2+DC2+2 AB·BD; but BC2=BD2 + DC2 (i. Prop. 47), and being substituted for them, AD2+DC=AB2+ BC2+2 AB BD. But AC2 = AD2 + DC2 (i. Prop. 47); and therefore AC2 AB2+BC2+2 AB BD. Consequently, AC2, or the square of the side subtending the

obtuse angle, exceeds AB2+ BC, or the sum of the squares of the sides containing that angle, by 2 AB BD, or twice the rectangle contained by the side AB and its external segment BD.

PROP. XIII. THEOR.

In any triangle (ABC) the square of a side (CB) subtending an acute angle, is less than the sum of the squares of the sides (CA, AB) containing that angle, by twice the rectangle contained by either of those sides (AB) and the part of it (AD) intercepted between the perpendicular (CD) let fall on it from the opposite angle and the vertex of the acute angle.

First, suppose the perpendicular CD to fall within the triangle ACB. Then, as AB is divided in D, AB2+AD2=

2 AB⚫AD + DB2 (ii. Prop. 7): add to both CD3, and then AB2+ AD2+ CD2= 2 AB·AD+DB2+ CD2. But AD2+ CD2= AC2 (i. Prop. 47), and DB2+ CD2= CB2. Therefore AB2+AC-2 AB AD+CB'; and consequently CB', or the square of the side subtending the acute angle, is less than AB2+AC2, or the sum of the squares of the sides containing

ር

that angle, by 2 AB AD, or twice the rectangle contained by one of them, and its segment adjacent to the acute angle. Next, suppose that the perpendicular CD falls outside of the triangle ABC. Then AD2+AB2=2 AD'AB +BD2 (ii. Prop. 7); add to both DC2, and then DC2 +AD2+AB2=2 AD AB+BD2+ DC2; but DC2 +AD2=AC2 (i. Prop. 47); and also BD2+DC2=BC2; and therefore AC2+AB2=2 AD AB+BC, as in the preceding case.

To draw a line (AB) of which the square shall be equal to a given rectilinear figure (F).

Construct the rectangular parallelogram DGHA equal to

E

H

B

the given figure F (i. Prop. 45); and if its sides G be equal it is a square, and any of its sides will solve the problem. But if its sides be not equal, produce one of them, as HA, till the produced part AE is equal to AD. Bisect HE in C, and from the centre C, with the interval CE, describe DA the semicircle HBE: produce DA till it meets the circumference in B. AB is the line required. For, join CB; and since HE is bisected in C, and divided unequally in A, HA⚫AE+CA2=CE2 (ii. Prop. 5): but CE2= CB2, since CE=CB (Def. 12); and CB2=AC2+ AB2 (i. Prop. 47). Therefore HA∙AE+CA2=CA2+AB2; and if CA2 be taken from both, there remains AB2= HA⚫AE= HA AD, since AE=AD (Const.). That is to say, the square of AB is equal to the rectangular parallelogram DGHA, which is equal to the figure F (Const.), and therefore the square of AB is also equal to the given figure (Ax. 1).

Note. See APPENDIX C.

[59]

BOOK III.

DEFINITIONS.

1. A RAY or radius of a circle is a straight line drawn from the centre to the circumference.

2. A CHORD is a straight line inscribed in a circle, and terminated both ways in the circumference.

3. A straight line is said to touch a circle when it meets, and being produced, does not cut the circle.

4. Circles are said to touch one another which meet but do no cut one another.

5. Straight lines are said to be equally distant from the centre of a circle when the perpendiculars drawn to them from the centre are equal.

6. And the straight line on which the greater perpendicular falls is said to be farther from the centre.

7. An ARCH or arc of a circle is any part of the circumference.

8. A SEGMENT of a circle is the figure contained by a straight line, and the arch which it cuts off.

1D

9. An angle in a segment is the angle made by two straight lines drawn from any point in the circumference of the segment to the extremities of the straight line, which is the base of the segment.

10. An angle in a segment is said to stand upon the arch intercepted between the straight lines which make the angle.

11. A SECTOR of a circle is the figure contained by two rays and the arch between them.

12. Similar segments of circles are those which contain equal angles.

PROP. I. PROB.

To find the centre of a given circle (AFB).

Draw any chord AB and bisect it in D (i. Prop. 10); draw DF perpendicular to AB (i. Prop. 11), and produce it till it meets the circumference in E. Bisect EF, and the point of bisection C is the centre of the circle.

B

D

E

For let it be assumed that any other point, as G, is the centre of the circle; join GD, GA, and GB; and then, since the triangles GDA, GDB have AD=DB (Const.), and DG common to both, and also GA = GB (Hyp. Def. 12), they must have the angles contained by the equal sides equal (i. Prop. 8); and therefore ▲ GDA = / GDB; and consequently GD must be perpendicular to AB (Def. 7), and GDB must be a right angle; but CDB is also a right angle (Const.); and therefore ▲ GDB = / CDB, a part equal to the whole, which is absurd. The point G, therefore, is not the centre of the circle; and the same may be demonstrated of every other point except C, which is therefore the centre.