SUMMARY OF PROPOSITION XLV., PROBLEM 13.

To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given angle.

Cons.-Divide given figure into triangles by joining BD. Make a parallelogram equal to one triangle, and having angle FKH equal to given angle (I. 42).

Apply to side GH a gram equal to the other triangle, and having an angle GHM or GLM equal to E (I. 44).

FKML is the gram required.

Proof.-FK and LM are both || to GH.

.. FK is to LM (I. 30).

▲ GHM =

FKH (cons.). Add KHG.

Zs GHM, KHG = ≤s FKH, KHG = 2 rt. angles

(I. 29).

.. KHM is a straight line (I. 14).

▲ FGH = ▲ FKH = ≤ GLM.
Ls FGH, HGL = Zs HGL,

(I. 29).

.. FGL is a straight line.

Add ▲ HGL.

GLM = 2 rt. angles

And since the parts of FL and KM are || (cons.), the whole lines are ||, and FKLM is a gram by construction, and it was made equal to the two parts of ABCD, and has an angle FKM equal to E.-Q.E.F.

LECTURE XVII.

THE SQUARE,-PROPOSITIONS 46-48.-ANALYSIS OF SECTION 3 OF BOOK I.

THE three concluding propositions of Book I. deal with the square. Proposition XLVI. requires us to describe a square on a given straight line AB. From def. 30 we learn that a square is a four-sided figure, having all its sides equal, and all its angles right angles. It is evident from this definition that, to begin with, we must make a right angle at A or B. How can we do this? Proposition XI. teaches us to draw a line at right angles to another from a given point in the same. From point A draw AC at right angles to AB; the angle BAC will form one of the angles of the square, and by cutting off from AC (produced if necessary) AD equal to AB we shall obtain another side. Since a square is also a parallelogram, the two remaining sides must be respectively parallel to AD, AB, already drawn. Draw through points B, D lines parallel to AD and AB, and meeting at a point E. Are the sides of this figure equal? ADEB is a parallelogram by construction, and its opposite sides are equal; therefore DE = AB, EB = AD, and AD was made equal to AB; consequently, the four sides are equal (ax. 1).

D

C

E

B

Are the angles all right angles ? < BAD is a right angle by construction; therefore, the opposite angle DEB is a right angle (I. 33). Again, because ED and AB are parallel, and AD meets them, the two interior angles EDA, DAB are equal to two right angles (I. 29), and BAD is a right angle (cons.); therefore, EDA is a right angle, and the opposite

angle ABE is a right angle (I. 33). That is, all the angles of ADEB are right angles, and all its sides are equal; therefore, it is a square, and it is described on the given line AB.

SUMMARY OF PROPOSITION XLVI., PROBLEM 14.

On a given straight line to describe a square.

Cons. From A draw AC at right angles to AB; cut off AD = AB.

Through B and D draw BE and DE || to AD, AB.

=

Proof.-ADEB is gram (cons.); .. DE

= AB, EB = AD.

AD AB (cons.); .. ADEB is equilateral.

LS BAD, ADE = 2 rt. angles (I. 29).

D

E

BAD is a rt. angle (cons.); .. 4 ADE is also.

Opposite angles of a gram are equal; .. remaining angles are rt. angles.

.. ADEB is a square, and it is described on AB.-Q.E.F.

Proposition XLVII. shows that, supposing squares to be described on all the three sides of a right-angled triangle, the squares on the sides containing the right angle are together equal in area to the square on the side opposite to the right angle. This side is frequently spoken of as the hypothenuse (VTOTEVOOα stretched under, i.e. the line stretched under the angle). Let us take any right-angled triangle BAC, and let BAC be the right angle. Having described squares on all three sides, we are required to show that the squares on BA, AC are together equal to the square on BC.

We may imagine that this theorem was discovered somewhat in the manner following.

First, assuming the theorem to be true, it must be possible to divide the square on BC into two parallelograms, which will be respectively equal to the squares on BA and AC. By I. 45 we can apply to one of the sides of the square on

BC, say BD, a parallelogram equal to one of the squares on the sides containing the right angle, say BA, and having one of its angles equal to BDE, i.e. a right angle. This parallelogram, which we will call BL, will, from its construction, coincide with a part of the square on BC, and the remaining part, the parallelogram LC, must be equal in area to the square on AC. Now, if you will look at the figure thus obtained, you will notice that the line defining the two grams will, if produced, pass through the point A. And you will find that this is the case in any right-angled triangle that can be drawn, providing that the constructed parallelogram be applied to that side of the square on the hypothenuse which is adjacent to the square to which the applied parallelogram is equal. This fact gives us the clue to a part of the construction required, for if the line which divides the square on BC into parallelograms equal to the squares on BA, AC always passes through A, it is easy to

construct the parallelograms required, by drawing through A a line parallel to BD or > CE cutting DE in L.

Now, adopting this construction, how can we show that BL is equal to the square on BA ? We know that parallelograms are equal under certain conditions when they are between the same parallels, but these are not between the same parallels. We know also

that parallelograms which are the complements of a larger parallelogram are equal, but these are not. What other test of equality have we, then? We have no direct test left, but we know that a parallelogram is double of a triangle on the same base and between the same parallels (1. 41). Then, if we can show that the square on BA, and BL are double of the same or equal triangles, the proof will be complete. Can

such a triangle or triangles be constructed? I think you will easily see that by joining AD we shall obtain a triangle ABD between the same parallels AL and BD, on same base BD as the parallelogram BL, and of which BL is therefore double.

But the square on BA and triangle ABD are obviously not between the same parallels, and we must endeavour then to construct a triangle similar to ABD and within the same parallels as the square on BA. Now triangle ABD is formed by a side of the square on BA and a side of the square on BC and the line AD joining their extremities. Then taking FB, BC and joining FC we shall obtain a triangle similar to ABD. About this triangle we must answer two questions. (1) Is it half of the square on BA? It is on the same base BF, and between the same parallels, for GA, AC are in the same straight line, since angles GAB, CAB are both right angles (I. 14). Therefore the square on AB is double of the triangle FBC (I. 41). (2) Is it equal to the triangle ABD ? sides FB, BA are equal, since they are sides of square on BA sides BC, BD are equal

BC

Angles ABD, FBC are each made up of a right angle+ angle ABC, and are therefore equal; and the triangle ABD is equal to the triangle FBC (I. 4). And since the parallelogram BL, and the square on BA, are respectively double of these triangles, they are equal.

In the same way by joining KB and AE we can show that the parallelogram CL is equal to the square on AC. Therefore the squares on AB, AC are together equal to the square on BC.—Q.E.D.

We proceed to sum up and arrange the synthesis of the proposition.

SUMMARY OF PROPOSITION XLVII., THEOREM 33.

In any right-angled triangle the square on the hypothenuse is equal to the squares on the sides containing the right angle. Cons. The squares on the three sides of triangle ABC being drawn,