Euclid, Elements of Geometry, Book I, Proposition 8
(Edited by Dionysius Lardner, 1855)

Proposition VIII. Theorem.
[Euclid, ed. Lardner, 1855, on Google Books]

(68) If two triangles (A B C and E F D) have two sides of the one respectively equal to two sides of the other (A B to E F and C B to D F), and also have the base (A C) equal to the base (E D), then the angles (B and F) contained by the equal sides are equal.

For if the equal bases A C, E D be conceived to be placed one upon the other, so that the triangles shall lie at the same side of them, A B C D E F and that the equal sides A B and E F, C B and D F be conterminous, the vertex B must fall on the vertex F; for to suppose them not coincident would contradict the seventh proposition. The sides B A and B C being therefore coincident with F E and F D, the angles B and F are equal.

(69)   It is evident that in this case all the angles and sides of the triangles are respectively equal each to each, and that the triangles themselves are equal. This appears immediately by the eighth axiom.

In order to remove from the threshold of the Elements a proposition so useless, and, to the younger students, so embarrassing as the seventh, it would be desirable that the eighth should be established independently of it. There are several ways in which this might be effected. The following proof seems liable to no objection, and establishes the eighth by the fifth.

Let the two equal bases be so applied one upon the other that the equal sides shall be conterminous, and that the triangles shall lie at opposite sides of them, and let a right line be conceived to be drawn joining the vertices.

1° Let this line intersect the base.

Let the vertex F fall at G, the side E F in the position A G, and D F in the position C G. A B C D E F G Hence B A and A G being equal, the angles G B A and B G A are equal (V). Also C B and C G being equal, the angles C G B and C B G are equal (V). Adding these equals to the former, the angles A B C and A G C are equal; that is, the angles E F D and A B C are equal.

2° Let the line G B fall outside the coincident bases.

The angles G B A and B G A, and also B G C and G B C are proved equal as before; A B C D E F G and taking the latter from the former, the remainders, which are the angles A G C and A B C, are equal, but A G C is the angle F.

3° Let the line B G pass through either extremity of the base.

In this case it follows immediately (V) that the angles A B C and A G C are equal; A B C D E F G for the lines B C and C G must coincide with B G, since each has two points upon it (52).

Hence in every case the angles B and F are equal.

This proposition is also sometimes demonstrated as follows.

Conceive the triangle E F D to be applied to A B C, as in Euclid's proof. A B C D E F Then because E F is equal to A B, the point F must be in the circumference of a circle described with A as centre, and A B as radius. And for the same reason, F must be on a circumference with the centre C, and the radius C B. The vertex must therefore be at the point where these circles meet. But the vertex B must be also at that point; wherefore &c.

Book I: Euclid, Book I (ed. Dionysius Lardner, 11th Edition, 1855)

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