But sq. on AB = sqq. on AD, BD. ·

Therefore the sq. on AC and twice the rect. CB, BD AB and BC.

=sqq.

This proof also applies to the second case, for though BD in Case I. corresponds to BC in Case II., the rect. CB, BD is contained by 'the whole line and a part' in each.

Euclid considers only the first case and enunciates the theorem 'In acute-angled triangles, &c.'

The converse of both of these propositions is easily demonstrated by the indirect method (as in I. 19).

SUMMARY OF PROPOSITION XII., THEOREM 11.

In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing it by twice the rectangle contained by one of these sides and the projection of the other upon it.

Sq. on AB sqq. on AC, CB, and twice rect. BC, CD.
Cons.-Nil.

Proof.-Sq. on AB = sqq. on AD, BD (I. 47).

Sq. on BD = sq. on BC, CD, and twice rect.

Sq. on AC

.. sq. on AB =

BC, CD (II. 4).

sqq. on AD, CD.

sqq. on AC, CB, and twice rect. BC, CD.-Q.E.D.

SUMMARY OF PROPOSITION XIII., THEOREM 12.

In acute-angled triangles the square on the side subtending the acute angle is less than the squares on the sides containing it by twice the rectangle contained by either of those sides and the projection of the other upon it.

Sq. on AC and twice rect. CB, BD = sqq. on AB, BC.
Cons.-Nil.

Proof.-Cases I. & II. Sq. on AC = sqq. on AD, DC (I. 47 ). Twice rect. CB, BD and sq. on DC sqq. on CB, BD (II. 7).

.. sq. on AC and twice rect. CB, BD

Case III. is manifest (I. 47).

=sqq. on CB, BD, DA.
=sqq. on CB, BA (I. 47).

Prop. XI., the first problem in the book is 'To divide a straight line into two parts, so that the rectangle contained by the whole line and one of the parts is equal to the square on the other.'

Adopting the method of analysis we will assume that the point X in the line AB is the one required. On AX describe the square AFGX, and producing GX to K, making XK = AB, construct the rectangle XBDK.

F

G

X

A

E

K

B

D

These figures are equal in area (hyp.); also AB and BD form two sides of the square on AB. Complete this square. Then the square on AB = the rectangle FK; since FX and XD are equal, and AK is common to both figures.

Now the rectangle FK is contained by a line CA produced to F and the part produced (AF = FG). Looking through the theorems on the segments of a straight line, we find that such a rectangle together with the square on half the original line the square on the line made up of half the line and the part produced. Bisect AC in E, then rect. CFFA and sq. on AE =sq. on EF (II. 6).

Also, since the sq. on AB = rect. CFFA,

Sq. on AB and sq. on AE = sq. on EF.

=

=

Now AB and AE are at right angles to each other, therefore joining their extremities we obtain a right-angled triangle, and the square on EB the hypothenuse the sqq. on AB, and AE sq. on EF; it follows then that EB = EF. And EF is made up of half AC AB, which is given, and AF a side of the square which we have assumed. EB then forms a measure for EF and enables us to determine the distance to which CA is to be produced. We can now retrace

=

our steps and obtain a construction which solves the problem

Sq. FH

On AB describe a square ABDC (I. 46).
Bisect AC in E. Join EB.

Produce EA to F, making EF = EB.
On AF describe sq. AFGH.

Produce GH to meet CD in K.

rect. HD, and H is the point required.

The proof is obvious from the analysis.

SUMMARY OF PROPOSITION XI., PROBLEM 1.

To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts shall be equal to the square on the other part.

Cons.-On AB describe the square ACDB (I. 46).

Bisect AC in E and join EB.

Produce CA to F, making EF = EB.

On AF describe the square AFGH.

Produce GH to K.

Proof-Rect. CFFA and sq. on AE sq. on EF (II. 6). =sq. on EB =sqq. on EA, AB.

.. rect. CFFA = sq. on AB.

Take from each the common part AK.

Then FH (sq. on AH) = HD (rect. AB, BH).

.. AH is divided at H, &c.-Q.E.F.

Prop. XIV. completes the problem of the quadrature of a rectilineal area, in which I. 42, 44, and 45 have been successive stages. It shows how to describe a square that shall be equal to a given rectilineal figure.' Let A be the given rectilineal figure, then, taking up the problem at the point we left off in I. 45, we can construct a parallelogram equal to A, having its sides inclined at any angle we please. Let them be inclined at right angles, and describe the rectangle BCDE equal to A. Then if BC= CD the rectangle is a square, and the problem is solved. But if not, it remains to construct a square equal to the rectangle BCDE.

A square and a rectangle were constructed equal in the figure of the last problem. Can we adopt that, or a similar construction ?

First, that the lines containing the rectangle may be represented by one straight line produce BE to F, making EF=ED. Then, looking through the theorems on the segments of a straight line, we find that if BF be bisected in G, the rectangle BE, EF (that is, the rectangle BCDE) with the square on GE=the square on GF (II. 5). In the last figure BE (=EF, the line corresponding to GF) is the hypothenuse of a right-angled triangle, of which the line corresponding to GE and a side of the required square contain the right angle. Can we construct such a triangle?

Since a line equal to GF is to be the hypothenuse, the vertex of the triangle must be in the circumference of a circle of which GF is a radius. Describe this circle, taking G as the centre.

Again, since GE is to be one of the sides containing the right angle, the vertex must be at some point in a line perpendicular to GE, that is in DE produced.

Thus either of the two points of intersection of DE produced with the circle will form the vertex of a triangle satisfying the required conditions.

Produce DE to meet the circumference of the circle in H and join GH.

Then HGE is a right-angled triangle and

sqq. on GE and EH=sq. on GH

Take

away

=

=sq. on GF=rect. BCDE & sq. on GE.

the common square and

sq. on EH=rect. BCDE

=given rectilineal figure A.

SUMMARY OF PROPOSITION XIV., PROBLEM 2.

To describe a square that shall be equal to a given rectilineal figure.

Cons.-Describe the rect. BCDE=A (I. 45).

Produce BE to F making EF=ED, bisect BF in G, with centre G and distance GF describe a semicircle.

Produce DE to meet it in H. Join GH.

=

Proof.-Rect. BE, EF and sq. on GE sq. on GF (II. 5) =sq. on GH=sqq. on GE and EH

.. rect. BE, EF alone = sq. on EH.
.. rect. BE, EF (i.e. BCDE=A)=sq. on EH.

Q.E.F.

The second book may be divided into two parts. Part I. consists of ten theorems showing the relations between the rectangles contained by various segments of a straight line. There is an analogy between these and certain elementary principles of algebra and arithmetic, thus II. 4 is the geometrical proof of the algebraical formula (a+b)2=a2+2ab+b2. The whole of Part I. may be demonstrated algebraically, but these demonstrations are not so satisfactory as Euclid's methods, owing to the existence of 'incommensurable magnitudes.' Thus by geometry we can express the exact length of the diagonal of a square which can only be determined approximately by algebra or arithmetic, e.g. let the side of a square be 1, 2, or 3 inches, then the diagonal is √2 √8 √18 in. and each of these numbers consists of an integer and an infinite decimal.

Part II. contains:-
:-

(a) Two theorems showing the relations between the squares on the sides of a triangle containing an acute or an obtuse angle, and the square on the side subtending it. These, with I. 47, are of great importance in trigonometry.

(8) Two problems, one showing how to divide a straight line into certain segments; the other, how to construct a