THEOREM (Euclid I. 5). Definition.-An isosceles triangle has two sides equal.

Repeat.-The enunciation of Euc. I. 4, and

Axiom 3.

General Enunciation.
The angles at the base of an isosceles tri-

angle are equal to one another; and if
the equal sides be produced the angles
on the other side of the base shall be

equal to one another. The same, in tabular form.

GIVEN.

REQUIRED.

(a) An isosceles triangle

To prove that (6) having the equal sides (a) The angles at the base produced.

are equal. (6) The angles on the

other side of the base

are equal. Particular Enunciation. Given.-The isosceles triangle (a) ABC, having

AB equal to AC, and the sides A B, AC

produced to D and E (Fig. 1). Required.-To prove that

(6) angle A B C is equal to angle A CB,

(c) angle D B C is equal to angle ECB. Construction.

In B D take any point F (Fig. 2). (d) From A E cut off a part A G' equal to AF

(by Euc. I. 3). Join B G and FC.

(The difficulty of this proposition arises from the fact thatthere are several triangles partly overlying each other.

A

B в

Fig. 1.

Thus we have to deal with two pairs of triangles. The two triangles ABG, ACF: and the two triangles BCF, CDG. You will understand this at once if you

cut out four triangles in pasteboard of the same shape as those in the figure, and lay them over one another as they are there.)

Proof.
Let us apply Euc. I. 4 to each of the pairs

of triangles.
First, we will take triangles ABG,ACF.

(In Fig. 3 those triangles are shown separately, as they would appear if lifted off one another.)

Fig. 3.

G

In these triangles we have

A F equal to AG (Construction d),
AC equal to AB (Given a).
angle FAC equal to angle GAB. (These angles

are equal because, in Fig. 2, where they lie one

on the other, they coincide.) .. by Euc. I. 4, we have

(e) base F C equal to base BG. (f) angle AFC equal to angle AGB. (8) angle ACF equal to angle A BG. Now, we take triangles B CF, CB G (Fig. 2). (In Fig. 4 those triangles are shown separately, as

they would appear if lifted off one another.
AF is equal to A G (by Construction d).

A B is equal to AC (Given a). (h).. The remainder BF is equal to the

remainder CG (Axiom 3).

Then, in the triangles B CF, CB G we have

BF equal to C G (h),
FC equal to GB (e),
and angle B F C equal to angle CGB (f).

by Euc. I. 4,
(i) angle FBC is equal to angle GCB, and

(k) angle FC B is equal to angle G B C. But it has been proved that

angle ABG is equal to angle A CF (8). (1) .. angle ABC is equal to angle ACB

(k and g). But, angles A B C, ACB (1) are the angles at the

base, and they are equal. And angles FBC, GCB (i) are the angles on the other side of the base, and they are equal.

Q. E. D. EXERCISE. - The pupil should now be asked to go through the proof, referring only to Fig. 2.

THEOREM (Euclid I. 6). Repeat.—The enunciation of Euc. I. 4 and

Axiom 9.

General Enunciation.
If two angles of a triangle be equal to one

another, the sides which are opposite to
the equal angles shall be equal to one

another. Particular Enunciation. Given.—The triangle ABC having (a) the

angle A B C equal to the angle ACB.

B

Required.–To prove that the side A B is equal

to the side A C. Construction.

Let us suppose that A B is not equal to A C;
Then one of the two must be greater.

Let us suppose that A B is greater than A C.
Let us also suppose (6) that D B is equal