PROP. X. (THEOREM.)—If a straight line (AB) be bisected (in C), and produced to any point (D), the squares of the whole line thus produced (AD), and of the part of it produced (DB), are together double of the square of half the line bisected (AC), and of the square of the line made up of the half and the part produced (CD).

E

Make

Through E

From the point C draw CE (I. 11) at right angles to AB. CE equal (I. 3) to AC or CB, and join AE and EB. draw EF parallel to AB (I. 31), and through D draw DF parallel to CE. Because the staight line EF meets the parallels CE, FD, the two angles CEF, EFD are equal (I. 29) to two right angles. Therefore the two 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

B

two right angles, will meet (Ax. 12) if produced far enough. Therefore EB and FD will meet if produced towards B and D. Let them meet in G, and join AG.

Because AC is equal to CE, the angle CEA is equal (I. 5) to the angle EAC. Because ACE is a right angle, each of the angles CEA and EAC is half a right angle (I. 32). For the same reason, each of the angles CEB and EBC is half a right angle. Therefore the whole angle AEB is a right angle. Because EBC is half a right angle, the vertical angle DBG is also (I. 15) half a right angle. But BDG is a right angle, being equal (I. 29) to the alternate angle DCE. Therefore the remaining angle DGB is half a right angle. Wherefore the angle DGB is equal to the angle DBG, and the side BD (I. 6) to the side DG. Again, because EGF is half a right angle, and the angle at F is a right angle, being equal (I. 34) to the opposite angle ECD, the remaining angle FEG is half a right angle. Therefore the angle FEG is equal to the angle EGF, and the side GF (I. 6) to the side FE. Because EC is equal to CA, the square of EC is equal to the square of CA. Therefore the squares of EC and CA are double of the square of CA. But the square of EA is equal (I. 47) to the squares of EC and CA. Therefore the square of EA is double of the square of AC. Again, because GF is equal to EF, the square of GF is equal to the square of EF. Therefore the squares of GF and FE are double of the square of EF. But the square of EG is equal (I 47) to the squares of GF and EF. Therefore the square of EG is double of the square of EF. But EF is equal (I. 34) to CD. Wherefore the square of EG is double of the square of CD. But it was proved that the square of EA is double of the square of AC. Therefore the squares of EA and EG are double of the squares of AC and CD. But the square of AG is equal (I. 47) to the squares of EA and EG. Therefore the square of AG is double of the squares of AC and CD. But the squares of AD and DG are equal to the square of AG. Therefore the squares of AD and DG are double of the squares of AC and CD. But DG is equal to DB. Therefore the squares of AD and DB are double of the squares of AC and CD. Wherefore, if a straight line, &c. Q. E. D.

PROP. XI. (PROBLEM.)-To divide a given straight line (AB) 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 (I. 46) the square CB. Bisect AC in E (I. 10), and join BE. Produce CA to F, and make EF equal (I. 3) to EB. Upon AF describe (I. 46) the square FH. The straight line AB is divided in H, so that the rectangle AB.BH is equal to the square of AH. Produce GH to meet CD in K.

A

E

F

C

a

B

K D

Because the straight line AC is bisected in E, and produced to F, the rectangle CF.FA together with the square of AE, is equal (II. 6) to the square of EF. But EF is equal to ER. Therefore the rectangle CF.FA, together with the square of AE, is equal to the square of E B. But the squares of BA and AE are equal (I. 47) to the square of EB. Therefore the rectangle CF.FA, together with the square of AE, is equal to the squares of BA and AE. From these equals take away the square of AE, which is common to both. Therefore the rectangle CF.FA is equal to the square of BA. But the rectangle FK is the rectangle contained by CF.FA, because FA is equal to FG. And AD is the square of AB. Therefore the rectangle FK is equal to the square AD. From these equals take away the common part AK. Therefore the remainder FH is equal to the remainder HD. But HD is the rectangle contained by AB.BH, because 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. Q. E. F.

PROP. XII. (THEOREM.)-In any obtuse-angled triangle (ABC), if a perpendicular be drawn from either of the acute angles (A) to the opposite side (BC) produced, the square of (A B) the side subtending the obtuse angle is greater than the squares of (AC and CB) the sides containing the obtuse angle, by twice the rectangle (BC. CD) contained by the side to which, when produced, the perpendicular is drawn, and the straight line intercepted between the perpendicular and the obtuse angle.

A

Because the straight line BD is divided into two parts at the point C, the square of BD is equal (11. 4) to the squares of BC and CD, and twice the rectangle BC. CD. To each of these equals add the square of DA. Therefore the squares of BD and DA are equal to the squares of BC, CD, and DA, and twice the rectangle BC. CD. But the square of BA is equal (I. 47) to the squares of BD and DA. And the square of CA is equal to the squares of CD and DA.

There

fore the square of BA is equal to the squares of BC and CA, and twice the rectangle BC.CD;

that is, the square of BA is greater than the squares of BC and CA by twice the rectangle BC. CD. Therefore in any obtuse-angled triangle, &c. Q. E D.

PROP. XIII. (THEOREM.)-In any triangle (ABC) the square of the side (AC) subtending either of the acute angles (B) is less than the squares of the sides (A B, BC) containing that angle, by twice the rectangle (CB, BD) contained by either of these sides, and the straight line intercepted between the acute angle and the perpendicular drawn to it from the opposite angle.

The square

To BB, one of the sides containing the acute angle B, draw the perpendicular AD from the opposite angle at A (I. 12). of AC, opposite to the angle B, is less than the squares of CB and BA by twice the rectangle CB.DB.

B D

A

B

C D

Because the straight line CB or BD is divided into two parts, at D or at C, the squares of CB and BD are equal (II. 7) to twice the rectangle CB.BD, and the square of DC. To each of these equals add the square of AD, Therefore the squares of CB, BD, and DA, are equal to twice the rectangle CB. BD, and the squares of AD and DC. But the square of AB is equal (I. 47) to the squares of BD and AD. And the square of AC is equal to the squares of AD and DC. Therefore the squares of CB and BA are equal to the square of AC, and twice the rectangle CB.BD; that is, the square of AC is less than the squares of CB and BA by twice the rectangle CB.BD.

When the perpendicular coincides with AC, the leg BC is the straight line between the perpendicular and the acute angle at B; and it is manifest that the squares of AB and BC are equal (I. 47) to the square of AC, and twice the square of BC. Therefore in any triangle, &c. Q. E. D.

R

PROP. XIV. (PROBLEM.)-To describe a square that shall be equal to a given rectilineal figure (A).

A

B

G

H

Describe the rectangular parallelogram BD equal (1. 45) to the rectilineal figure A. If the sides BE and ED of the rectangle BD are equal to one another, it is a square, and what was required is done. But, if they are not equal, produce one of them BE to F, and make (I. 3) EF equal to ED. Bisect (I. 10) BF in G. From the centre G, at the distance GB or GF, describe the semicircle BHF. Produce DE to meet the circumference in H. The square described upon EH is equal to the given rectilineal figure A. Join GH.

Because the straight line BF is divided into two equal parts at G, and into two unequal parts at E; the rectangle BE EF, together with the square of EG, is equal (II. 5) to the square of GF. But GF is equal (Def 15) to GH. Therefore the rectangle BE. EF, together with the square of EG, is equal to the square of GH. But the squares of HE and EG are equal (I. 47) to the square of GH.

D

Therefore the rectangle BE. EF, together with the square of EG, is equal to the squares of HE and EG. From these equals take away the square of EG, which is common to both. Therefore the rectangle BEEF is equal to the square of HE. But the retangle contained by BE. EF is the parallelogram BD, because EF is equal to ED. Therefore BD is equal to the square of EH. But BD is equal (Const.) to the rectilineal figure A. Therefore the square of EH is equal to the rectilineal figure A. Wherefore the square described upon EH, is equal to the given rectilineal figure A. Q. E. F.

BOOK III.

DEFINITIONS.

I.

EQUAL circles are those of which the diameters are equal, or those from the centres of which the straight lines drawn to the circumferences are equal.

"This is not a definition, but a theorem, the truth of which is evident; for, if the circles be applied one to another, so that their centres coincide, the circles must likewise coincide, since the straight lines drawn from the centres are equal."

II.

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

III.

Circles are said to touch one another when their circumferences meet in any point, but do not cut one another.

IV.

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.

V.

And the straight line which has the greater perpendicular drawn to it is said to be farther from the centre.

VI.

A segment of a circle is the figure contained by a straight line or chord, and the arc, or part of the circumference which cuts it off.

VII.

The angle of a segment is that which is contained by the straight line and the circumference.

VIII.

An angle in a segment is the angle contained 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.

IX.

An angle is said to insist or stand upon the circumference intercepted between the straight lines that contain the angle.

X.

A sector of a circle is the figure contained by two (radii, or) straight lines drawn from the centre, and the arc, or part of the circumference between them.

XI.

Similar segments of circles are those in which the angles are equal, or which contain equal angle.

PROP. I. (PROBLEM.)-To find the centre of a given circle (ABC). Take any two points A and B, in the circumference, and join AB. Bisect the straight line AB (I. 10) at D. From the point D draw DC at right angles, (I. 11) to AB. Let CD meet the circumference in C and E; and bisect C E in F. The point Fis the centre of the circle ABC. For if it be not, let, if possible, G, a point not in CE, be the centre, and join GA, GD and GB.

C

FG

E

D

B

Because DA is equal (Const.) to DB, and DG common to the two triangles ADG, BDG, the two sides A D, DG, are equal to the two sides BD, DG, each to each. But the base GA is equal (I. Def. A 15) to the base GB, because they are drawn from the centre G. Therefore the angle ADG is equal (I. 8) to the angle GDB. But when a straight line standing upon another straight line makes the adjacent angles equal to one another, each of these angles is a right angle (I. Def. 10). Therefore the angle GDB is a right angle. But FDB is likewise (Const.) a right angle. Wherefore the angle FDB is equal (I. Ax. 1) to the angle GDB; that is, the greater equal to the less, which is impossible. Therefore G is not the centre of the circle ABC. In the same manner it can be shown that no other point out of CE is the centre. Because CE is bisected in F, any other point in CE divides it into unequal parts, and cannot be the centre. Therefore no other point but F can be the centre. Wherefore F is the centre of the circle ABC. Q. E. F.

Cor. From this it is manifest, that if in a circle a straight line bisects another at right angles, the centre of the circle is in the line which bisects the other.

PROP. LI. (THEOREM.)—If any two points (A and B) be taken in the circumference of a circle (ABC), the straight line which joins them lies within the circle.

Find D the centre of the circle ABC (III. 1), and join DA and DB. In the arc AB, take any point F; join DF, and let it meet the straight line AB in É.