Through A draw AL to BD or CE and meeting BD

art. angle + ABC)

sq. on BA is double of FBC.

||gram BL is double of ▲ ABD.

=

.. sq. on BA ||gram BL.

In same manner show that LC = sq. on AC, &c.

The discovery of this theorem is ascribed to Pythagoras; it is said that he was so delighted at his success that he sacrificed an ox in honour of the occasion.

I. 48. The book concludes with the converse of this proposition, which, however, is not demonstrated indirectly, as is so frequently the case. In his proof Euclid makes use of two facts which are evident from the nature of the square, but to which he has not previously called attention: (1) that squares on equal straight lines are equal, and (2) conversely that equal squares must be on equal straight lines. The theorem to be proved is thus enunciated: if the square described on one of

the sides of a triangle be equal to the squares described on the other two sides of it, the angle contained by these two sides is a right angle.

Let BAC be a triangle such that the squares on BA, AC are together equal to the square on BC, it follows that the angle BAC is a right angle.

Euclid proves this by constructing on AC a triangle similar to BAC. This is done by drawing a line at right angles to AC, making a part AD equal to AB and joining DC.

B

D

Then since angle DAC is a right angle by construction, squares on DA, AC =square on DC.

And since DA = AB (cons.), squares on DA, AC = squares on BA, AC square on BC (hyp.).

=

Therefore the square on DC = square on BC, and line DC =line BC.

Then by I. 8 the triangles DAC and BAC are equal, and the angle BAC = angle DAC.

But DAC is a right angle (cons.), therefore the angle BAC is a right angle.-Q.E.D.

SUMMARY OF PROPOSITION XLVIII., THEOREM 34.

If the squares on the sides containing an angle of a triangle are together equal to the square on the line subtending it, the angle in question is a right angle.

AB.

Cons.-Draw a line at right angles to AC, make AD =

Join DC.

Proof.-Sqq. DA, AC sqq. on BA, AC

The third section of Book I. consists of:

I. Eight theorems giving the conditions of equivalence of area in triangles and parallelograms (Props. 35–41, and 43). The results of these have been already tabulated.

II. Three problems applying these conditions, viz. to describe a parallelogram equal (a) to a given triangle, with or without one side being given, (B) to any given rectilineal figure. In all three problems one angle of the parallelogram required is to be equal to a given angle. (I. 42, 44, 45.)

III. Three propositions, one problem, and two theorems relating to the square.

On a given straight line to describe a square (I. 46).

The square on the hypothenuse is equal to the squares on the sides containing the right angle, in any right-angled triangle, and conversely (I. 47, I. 48).

EUCLID'S ELEMENTS.

BOOK I.

PROPOSITION I., PROBLEM 1.

To describe an equilateral triangle on a given finite straight

line.

Let AB be the given straight line:

It is required to describe an equilat. A on AB.

From the centre A, at the

distance AB, describe the circle BCD. (Post. 3.)

From the centre B, at the distance BA, describe the circle ACE. (Post. 3.)

To the point C at which the

circles cut one another, from

D

A

points A and B draw the straight lines AC, BC. (Post. 1.) ABC shall be an equilateral triangle.

For. A is the centre of the circle BCD, AC

= AB

And B is the centre of the circle ACE, BC = AB

.. AC, BC each = AB.

But things which are equal to the same thing are equal to one another (ax. 1).

Thus AC, AB, BC are all equal, and an equilat. ▲ ABC has been described on AB.-Q.E.F.

PROPOSITION II., PROBLEM 2.

From a given point to draw a straight line equal to a given straight line.

Let A be the given point, and BC the given straight line it is required to draw from A a straight line equal to BC.

;

Join A, B. (Post. 1.) On AB describe the equilateral A DAB (I. 1). Produce the straight lines DA, DB to E and F. (Post. 2.)

From the centre B, at distance BC, describe the circle CGH meeting DF in G. (Post. 3.)

From the centre D, at

the distance DG, describe the circle GKL meeting DE in L.

15).

For
And

AL shall be equal to BC

B is the centre of circle CGH, BC = BG

D is the centre of circle GKL, DL = DG. (def.

Also parts of these DA and DB are equal (def. 24). .. remainder AL= remainder BG (ax. 3).

But BC= BG; .. AL= BC.

Thus from point A a straight line AL has been drawn= BC.-Q.E.F.

PROPOSITION III., PROBLEM 3.

From the greater of two given straight lines to cut off a part equal to the less.

Let AB and CD be the two given straight lines, of which AB is the greater.

It is required to cut off from AB a part = CD.