The following is as given in
Sir Thomas L. Heath's translation, which can be found in the book
The Thirteen Books of The Elements, Vol. 1. This well-known demonstration is known in Latin as the *pons asinorum* or "bridge of fools", possibly because the construction looks like a bridge, and because, as this is the first proposition in the Elements to require a fair bit of thought to prove, it may have been used in antiquity to separate those knowledgeable in geometry from those who are not.

Book I | Book II | Book IX |
---|---|---|

Definitions, Postulates, and Common Notions | Definitions | Proposition 20 |

Proposition 1, Proposition 3, | Proposition 14 | Proposition 36 |

Proposition 5, Proposition 6, | ||

Proposition 29, Proposition 47 |

*In isosceles triangles the angles at the base are equal to one another, and, if the equal straight lines be produced further, the angles under the base will be equal to one another.*

Let `ABC` be an isosceles triangle having the side `AB` equal to the side `AC`;

and let the straight lines `BD`, `CE` be produced further in a straight line with `AB`, `AC`. [Post. 2]

I say that the angle `ABC` is equal to the angle `ACB`, and the angle `CBD` to the angle `BCE`.

Let a point `F` be taken at random on `BD`;

from `AE` the greater let `AG` be cut off equal to `AF` the less; [I. 3]

Then, since `AF` is equal to `AG` and `AB` to `AC`,

the two sides `FA`, `AC` are equal to the two sides `GA`, `AB`, respectively;

and they contain a common angle, the angle `FAG`.

Therefore the base `FC` is equal to the base `GB`,

and the angle `AFC` to the angle `AGB`. [I. 4]

And, since the whole `AF` is equal to the whole `AG`,

and in these `AB` is equal to `AC`,

the remainder `BF` is equal to the remainder `CG`.

But `FC` was also proved equal to `GB`;

therefore the two sides `BF`, `FC` are equal to the two sides `CG`, `GB` respectively;

and the angle `BFC` is equal to the angle `CGB`,

while the base `BC` is common to them;

therefore the triangle
therefore the angle `FBC` is equal to the angle `GCB`,

and the angle`BCF` to the angle `CBG`.

and the angle

Accordingly, since the whole angle `ABG` was proved equal to the angle `ACF`,

and in these the angle `CBG` is equal to the angle `BCF`,

the remaining angle
and they are at the base of the triangle `ABC`.

But the angle`FBC` was also proved equal to the angle `GCB`;

and they are under the base.

But the angle

and they are under the base.

Therefore etc. Q.E.D.