EF is double of the square of GF. But GF is equal (I. 34) to CD. Therefore the square of EF is double of the square of CD. But it has been proved that the square of AE is double of the square of AC. Therefore the squares of AE and EF are double of the squares of AC and CD. But the square of AF is equal (I. 47) to the squares of AE and EF. Therefore the square of AF is double of the squares of AC and CD. But the squares of AD and DF are equal (I. 47) to the square of AF. Therefore the squares of AD and DF are double of the squares of AC and CD. But DF is equal to D B. Therefore the squares of AD and DB are double of the squares of AC and CD. If, therefore, a straight line be divided, &c. Q. E. D.

Otherwise.-Because B C is divided into any two parts at D, the square of B D, and twice the rectangle B C. CD are together equal (II. 7) to the squares of BC and CD. But AC is equal (Const.) to B C. Therefore, the square of BD and twice the rectangle A C. CD are together equal to the squares of AC and CD. But the square of AD is equal (II. 4) to the squares of AC and CD, and twice the rectangle A C. CD. Therefore adding these equals, the squares of A D and BD, and twice the rectangle A C. CD are together equal (I. Ax. 2) to twice the squares of A C and C D, and twice the rectangle A C. CD. From these equals, take away twice the rectangle A C. CD. Therefore the squares of AD and BD are equal (I. Ax. 3) to twice the squares of A C and CD.

PROP. X. THEOREM.

If u straight line be bisected, and produced to any point, the squares of the whole line thus produced, and 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 and DB, are together double of the squares of AC and CD.

E

B

F

G

From the point C draw CE (I. 11) at right angles to AB. Make CE equal (I. 3) to A Cor CB, and join AE and EB. Through E draw EF parallel to AB (I. 31), and through D draw DF parallel to CE. Because the straight 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 two right angles, will meet (Ax. 12) if produced far enough. Therefore E B and FD will meet, if produced towards B and D. Let them meet in G, and join A G.

Because A C is equal to CE, the angle CEA is equal (I. 5) to the angle EA C. Because ACE is a right angle, each of the angles CEA and E AC 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 A EB 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

E

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 E C is equal to CA, the square of E C 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 E A is equal (I. 47) to the squares of EC and CA. Therefore the square of E A is double of the square of A C. Again, because GF is equal to EF, the square of G F 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 E G 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 E G 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 A G is equal (I. 47) to the squares of EA and EG. Therefore the square of A G 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.

Otherwise.-Produce CA, making CH equal (I. s) to CD. To these equals, add
the equals CB and CA, respectively.
Therefore HB is equal (I. Ax. 2) to AD.
Because the squares of H B and BD are

together (II. 9) double of the squares

H A

C

B D

of A C and CD. But HB is equal to AD. Therefore the squares of AD and BD are together double of the squares of A C and CD together.

PROP. XI. PROBLEM.

To divide a given straight line into two parts, so that the rectangle contamed

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 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 A C 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

G

C

EB

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 E F. But EF is equal to EB. Therefore the rectangle CF. FA, together with the square of AE, is equal to the square EB. But the squares of B A 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 B A. 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 A H. Q. E. F.

Dr. Thomson in his edition, has made an improvement in the construction of this problem, by cutting off at once from AB, a part AH equal to AF, and then completing the square F H for the demonstration. This removes the hiatus in the demonstration of Euclid, which does not show that AH and AB, the sides of the squares F H and AD, must coincide. It creates another, however, for it is not shown by Dr. Thomson that A B is the greater of the two straight lines A B and AF. But the same defect lurks in the demonstration of Euclid. Corollary 1.-In the figure to this proposition, the straight line CF is cut at the point A, similarly to the straight line A B at the point H.

Corollary 2.-If a straight line be divided, so that the rectangle contained by the whole and the smaller segment is equal to the square of the greater, the greater segment will be similarly divided by cutting off from it a part equal to the smaller segment; and the smaller segment will be similarly divided by cutting off from it a part equal to their difference, and so on continually.

PROP. XII. THEOREM.

In obtuse-angled triangles, if a perpendicular be drawn from either 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 to which, when produced, the perpendicular is drawn, and the straight line intercepted 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 perpendicular to BC produced. The square of AB is greater than the squares of A C and CB, by twice the rectangle B C. CD.

A

Because the straight line BD is divided into two parts at the point C, the square of BD is equal (II. 4) to the squares of B C and CD, and twice the rectangle BC. CD. To each of these equals, add the square of D A. Therefore the squares of BD and DA are equal to the squares of B C, CD and D A, and twice the rectangle B C.C D. But the square of BA is equal (I. 47) to the squares of BD and D A. And the square of CA is equal to the squares of CD and DA. Therefore the square of BA is equal to the squares of B B C and C A, and twice the rectangle B C. CD; that is, the square of BA is greater then the squares of BC and CA, by twice the rectangle BC.CD. Therefore, in obtuse-angled triangles, &c. Q. E. D.

C

D

PROP. XIII. THEOREM.

T every triangle, the square of the side subtending either 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 acute angle and the perpendicular drawn to it from the opposite angle.

Let ABC be any triangle, and the angle at B one of its acute angles; and to B C, one of the sides containing it, draw the perpendicular AD from the opposite angle at A (I. 12). The square of AC opposite to the angle B, is less than the squares of CB and B A by twice the rectangle C B.D B.

B D

A

C

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 DA. And the square of AC is equal to the squares of AD and D C. Therefore the squares of CB and BA are equal to the square of A C, 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 B C, are equal (I. 47) to the square of AC, and twice the square of BC. Therefore in any triangle, &c. Q. E 1).

Corollary.-The square of any side of a triangle is greater than equal to, or less than the squares of the other two sides according as the angle opposite to that side is greater than, equal to or less than a right angle; and the difference, where it exists, is twice the rectangle con- B C tained by either of the other sides, and the straight line intercepted

between the vertex of that angle and a perpendicular drawn to the remaining side from its opposite angle.

No 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.

Describe the rectangular parallelogram BD equal (I. 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

G

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 cireumference in H. The square described upon EH is equal to the given rectilineal figure A. Join G H.

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. Therefore the rectangle BE. EF, together with the square of EG, is equal to the squares of HE and E G. From these equals, take away the square of EG, which is common to both. Therefore the rectangle BE. EF is equal to the square of HE. But the rectangle 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.

Corollary.-The square of a perpendicular drawn from any point in a circle to its diameter is equal to the rectangle contained by the segments into which it divides the diameter.

The following Propositions and Corollaries are added to this Book in some editions of Euclid. As they are comparatively easy to the student who has mastered the first and second books, we give them as exercises.

PROP. A. THEOREM.-The squares of any two sides of a triangle are together double of the squares of half the third side and of the straight line drawn from the opposite angle bisecting that side.

PROP. B. THEOREM.-The squares of the two diagonals of a parallelogram are together equal to the squares of its four sides.

PROP. C. THEOREM.-The squares of the four sides of a trapezium are together equal to the squares of its two diagonals, and four times the square of the straight line which joins the points of the bisection of the diagonals.

PROP. D. PROBLEM.-To divide a given straight line into two parts so that the rectangle contained by its segments shall be equal to a given square, not greater thau the square of half the given straight line.

PROP. E. PROBLEM.-To produce a given straight line, so that the rectangle contained by the whole line thus produced and the part produced, shall be equal to a given square.