CASE II.-When the angle BCA is a right angle.

BC is now the same as BD, and the square on BC is the rectangle

BC, BD.

By I. 47 the square on AC is less than the sum of the squares on BC, BA by twice the square on BC.

Q. E. D.

Prop. 12 may be expressed thus :—

(1) AB2 = BC2 + CA2 + 2BC.CD;

=

a2 + b2 + 2ad, where a, b, c denote the sides of the triangle ABC, and d denotes CD;

or (2) c2

(1) CA2 or (2) 62

=

2BC.BD;

a2 + c2 2ad, where a, b, c denote the sides of the triangle ABC and d denotes BD;

or (3) 62 = a2 + c2 2ac cos B, for BD = c cos B trigonometrically. Since BD is the projection of BA on BC (see p. 100), Prop. 13 may be enunciated as follows :-In every triangle the square on the side subtending an acute angle, is less than the sum of the squares on the sides containing that angle, by twice the rectangle contained by either of these sides and the projection of the other side upon it.

EXERCISES ON PROPOSITIONS 9 AND 10.

1. In the figure of Prop. 10 produce BA to R, making RA equal to BQ, and then deduce Prop. 10 from Prop. 9.

2. In AB the diameter of a circle, any two points C and D are taken equally distant from the centre, and any point P is taken on the circumference; show that the sum of the squares on PC, PD will be equal to the sum of the squares on AC, AD.

3. If the straight line AB be divided at C so that the square on AC is double the rectangle AB, CB, prove that the sum of the squares on AB, CB is double the square on AC.

4. Show that Props. 9 and 10 are both included in the theorem :-The sum of the squares on the sum and the difference of two straight lines is equal to twice the sum of the squares on the lines.

EXERCISES ON PROPOSITIONS 11, 12, 13.

1. In the figure of Prop. 11, prove that

(i) CF is divided in medial section at A.

(ii) CH produced and FB intersect at right angles.
(iii) the ratio of AH to HB is that of √5 1 to 3
(iv) the lines AK, FD and GB are parallel.

(v) if CH and EB meet in N, AN is perpendicular to CH.

No5.

(vi) the square on EF is equal to five times the square on CE.

(vii) if BG is bisected at M, MH is equal to MB.

(viii) the sum of the squares on AB, HB is equal to three times the

square on AH.

(ix) that AH is greater than HB, but less than twice HB.

(x) the difference of the squares on AB, AH is equal to the rectangle

AB, AH.

(xi) the square on a line equal to the sum of AB and HB is equal to five times the square on AH.

2. Enunciate Prop. 12, using the projection of one side on another.

3. Taking the three figures in Prop. 13, write down the values of AB2, of BC2, and of CA2 in each, and express the results algebraically and trigonometrically in each case.

4. Two right-angled triangles ACB, ADB stand on the same hypotenuse AB and on the same side of it; if AD and BC intersect in 0, show that the rectangle AO, OD is equal to the rectangle BO, OC.

5. The sides of a triangle are four, five, and seven inches respectively; find the nature of each of its angles and also the length of the projection of the side four inches long upon the side five inches long.

6. State and prove the converse of Prop. 13.

EUC.

L

PROP. 14.-Problem.-To 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 equal to A.

Construction. Describe the parallelogram BCDE equal to the rectilineal figure A and having the angle CBE a right

Then if BC is equal to BE, BD is a square, and what was required

is done.

But if BC is not equal to BE, produce BE to F, making EF equal

bisect BF at G, and with centre G and radius GF describe the

semi-circle BHF;

produce DE to meet the circumference in H.

Then the square on EH shall be equal to the given figure A.

Join GH.

Proof. Because BF is divided equally at G and unequally at E, therefore by Prop. 5 the rectangle BE, EF, together with the square on GE, is equal to the square on GF;

I. 47

but the square on GF is equal to the square on GH, and the square on GH is equal to the sum of the squares on GE, EH, for the angle GEH is a right angle; therefore the rectangle BE, EF, together with the square on GE, is equal to the sum of the squares on GE, EH;

take away the common square on GE,

therefore the rectangle BE, EF is equal to the square on EH. But the rectangle BE, EF is equal to BD, for ED is equal to EF; and BD is equal to A;

therefore the square on EH is equal to the given figure A.

Q. E. F.

NOTES ON THE PROPOSITIONS IN BOOK II.

The propositions in the second book of Euclid are often found to be the hardest and the most uninteresting in all Euclid. This applies more especially to the first seven propositions in the book.

The Method of Proof used by Euclid in all these seven propositions depends on the fact that the area of any rectilineal figure is equal to the sum of the areas of all the separate parts of the figure. In other words we look at a certain figure and our eyes tell us that it is equal to certain other figures. The difficulty arises when we come to express this equality in Euclidean language. The chief difficulty, therefore, in Book II. lies in remembering and distinguishing the enunciations of the propositions, and for this reason the student, when repeating an enunciation in this book, should also state the proposition in letters, as given after each proposition, e. g.–

Proposition 1.-This is the only proposition in Book II. dealing with two lines. From it Props. 2 and 3 can be deduced, and from these all the propositions up to Prop. 10 can successively be deduced. Prop. 1 may, therefore, be considered the fundamental proposition of Book II. The proof of Prop. 1 depends on the evident fact that the rectangle AG is equal to the sum of the rectangles AH, CK, and DG. Then each of these rectangles is taken in order and changed into an equal rectangle.

Proposition 2.-This proposition is a special case of Prop. 1 obtained by making the line X equal to AB. The proof depends on the fact that AE is equal to the sum of AF and CE, and then each of these figures is changed in order into an equal square or rectangle.

Proposition 3.-This proposition is a special case of Prop. 1 obtained by making the line X equal to AC. The proof depends on AF being equal to its parts AE and CF, and these are changed into an equal rectangle, square, and rectangle respectively.

Propositions 4, 5, 6, 7.-The proofs given first for these propositions are Euclid's proofs; the alternative proofs are obtained from the previous propositions in Book II. Euclid's proofs of these four propositions are all based on the same method, and, like the proofs of Props. 1, 2, 3, they all depend on a certain figure being equal to the sum of its parts and on these parts being equal to certain other rectangles or squares.

The following points about Euclid's proofs of Props. 4, 5, 6, 7 may help the student

(1) All the proofs begin with "Because the complement, etc."

(2) All the proofs aim first at getting the required rectangle, 2 AC.CB, or AQ.QB, or 2 AB. CB.

(3) (a) In Prop. 4 nothing is added to the complements.

(b) In Prop. 5 QL and PF are added.

(c) In Prop. 6 PL is added.

(d) In Prop. 7 CK is added at once to both complements.

(4) Props. 5 and 6 are two cases of one proposition, the point Q being either internal or external to AB; the rectangle is the same in both.

(See Ex. 8, p. 139.)

(5) In Prop. 7 the letters AB.CB are repeated for squares and for the

rectangle.

The Alternative proofs of Props. 4, 5, 6, 7 have the advantage of being much shorter than Euclid's proofs; and, being based on former propositions of Euclid, they are perfectly satisfactory. The student need not hesitate therefore to adopt these proofs; but Euclid's proofs are useful as illustrations of a different method of proof, and to the mathematician the method of proof employed is often of more interest than the actual result obtained.

The following points about these Alternative proofs may be noticed :(1) All begin with getting the first side of the required equation; e. g. Props. 4 and 7 begin with the square on AB, and Props. 5 and 6 with the rectangle AQ.QB.

(2) The proofs of 5 and 6 are really the same, allowance being made for the different positions of Q, either internal or external to AB.

(3) All the proofs end with an application of Prop. 3.

N. B.-In Prop. 6 the rectangle is given as AQ, QB because it occurs in the same form in Prop. 5, and it is of great importance that the student should from the first recognize the similarity of Props. 5 and 6. The same collocation of letters, AQ, QB, also occurs in Props. 9 and 10, and this will help the beginner to remember the propositions. The student can afterwards easily alter the rectangle AQ, QB into AQ, BQ in Prop. 6, if he wishes to be strictly accurate. The latter form is the more correct, because, as already explained, QB in Prop. 6 is to be considered negative and BQ positive. (See p. 118.)