IT IS REQUIRED TO PROVE that the triangles are equal in all respects.

Apply the triangle ABC to the triangle DEF, so that B may be on E, and BA on ED, and C on the same side of ED as F.

Because the angle ABC is equal to the angle DEF,

Therefore BC falls on EF

Hyp.

If A does not fall on D, nor C on F, but A is in DE and C in EF, or A in ED produced, and C in EF produced, the impossibility which can be shown is that the whole is equal to its part.

If A is in ED, and C in EF produced, or if A is in ED produced, and C in EF, the impossibility which can be shown is that an exterior angle is equal to an interior and opposite angle. (In the above diagram the angle A'C'E would be both equal to and greater than the angle DFE.) Because an impossibility follows if A is not on D, nor C on F, Proved. Therefore either A is on D, or C is on F,

and the triangles are equal in all respects.

(b) In the triangles ABC and DEF;

I. 26.

GIVEN that the side AB is equal to the side DE, that the angle ABC

is equal to the angle DEF, and that the areas are equal,

IT IS REQUIRED TO PROVE that the triangles are equal in all respects.

B

A

A

E

Apply the triangle ABC to the triangle DEF, so that ▲ may be on D, and AB on DE, and C on the same side of DE as F.

Because AB is equal to DE,

Therefore B falls on E.

Because AB falls on DE,

and the angle ABC is equal to the angle DEF,

Therefore BC falls on EF.

Because if C does not fall on F, the whole can be shown
equal to its part, which is impossible,

Therefore C falls on F, and the triangle ABC is equal
to the triangle DEF in all respects.

Нур.

Proved.

Нур.

Ax. 9.

I. 4.

The other case of (b), viz. when the given side is opposite to the given angle can be best demonstrated after the properties of the circle have been discussed.

(c) is included in (a), and (d) in (b).

(151) The straight line drawn from the vertex of any triangle to the middle point of the base, bisects the triangle.

(152) If the four triangles into which the diagonals of a quadri ́lateral divide it are equal in area, the quadrilateral is a parallelogram.

(153) The four triangles into which a parallelogram is divided by its diagonals, are equal in area.

(154) The straight line joining the middle points of two sides of a triangle is parallel to the other side.

(155) The straight lines which join the middle points of the three sides of a triangle, divide the triangle into four equal parts.

(156) If a quadrilateral having two of its opposite sides parallel is bisected by one of its diagonals, it is a parallelogram.

(157) Any straight line drawn through the middle point of a diagonal of a parallelogram, bisects the parallelogram.

P. G.

4

The student should now read the remaining six propositions of the first book.

I. 43. In the forty-third proposition, as in the thirty-fifth, it is to be noticed that the remainders-in this case the complements--although equal in area, are not necessarily capable of being superposed so as to coincide.

The student should determine under what conditions the complements of a parallelogram are equal in all respects.

I. 45. In the forty-fifth proposition the rectilineal figure may have any number of sides, and can always be divided into triangles; there being in any case two less triangles than sides. There are therefore two less parallelograms to be described than there are sides in the given figure.

I. 47. The forty-seventh proposition is a very important one, and a great many demonstrations of it have been given. One of these demonstrations shows how the two similar squares can be cut in pieces and fitted together so as to form the larger square.

It should be noticed that Euclid, without proof, assumes that BL is a parallelogram.

I. 48. Before the forty-eighth proposition, two additional propositions ought to have been proved, viz:-Equal squares are upon equal straight lines, and, conversely, The squares on equal straight lines are equal.

In the forty-seventh and forty-eighth propositions we have an instance of a proposition and its converse being both proved directly. The forty-eighth proposition might equally well have been proved indirectly, if the constructed triangle had been drawn on the same side of the line AC as the other triangle.

If the proposition "If A is equal to B, then C is equal to D" and its obverse "If A is not equal to B then C is not equal to D" be both true, then we know that the converse, viz. "If C is equal to D then A is equal to B" is also true (see page 10).

The following propositions of Euclid's first book, afford examples of the above principle.

(158) The square on the diagonal of a square is double of the square itself.

(159) Construct a rectangular parallelogram, having two adjacent sides equal to two given straight lines.

(160) Describe a square equal to the sum of two given squares. (161) Describe a square equal to the difference of two given

squares.

(162) In I. 47 the straight lines BG and CH are parallel, and the four triangles ABC, GAH, KCE and FBD are equal.

(163) If the perpendicular from the vertex of a triangle divides the base into two segments the difference of the squares on these segments is equal to the difference of the squares on the sides.

CHAPTER IV.

In the fourth and eighth propositions Euclid superposes the triangles, and concludes their equality by showing that they can be made to coincide.

He might also have used the method of superposition to demonstrate his twenty-sixth proposition, case 1 directly, and case 2 indirectly.

The same figure can be imagined in two different positions, and then the one concept placed on the other.

A triangle ABC can be imagined in another position as A'B'C', and then ABC applied to A'B'C' so as to test the equality of the sides or angles. In this way propositions five and six can be demonstrated.

Superposition can prove inequality as well as equality, proposition 24 can be demonstrated by means of it.

The following theorems are given as illustrations of the application of the method of superposition.