Bridge of Asses or Pons Asinorum (Latin for "Bridge of Asses") is a term used to refer to a problem that severely tests the ability of an inexperienced person, and therefore separates the serious and dedicated students from the “asses.” It is said that students are as reluctant to tackle these problems as donkeys (asses) are to cross over a bridge. Once a student is experienced in his field, however, the problem appears relatively simple. The term can be used to refer to a problem that is a stumbling block in any field, or to a problem whose solution seems pointless.
The term “Bridge of Asses” first came into use during the Middle Ages, and is most commonly applied to a diagram used to help students of logic identify the middle term in a syllogism, or to Euclid's fifth proposition in Book 1 of his Elements of geometry. As early as the sixth century, the Greek philosopher Philoponus used a diagram to show what kind of conclusions (universal affirmative, universal negative, particular affirmative, or particular negative) follow from what kind of premises.
The sixth century Greek philosopher Philoponus, presented a diagram showing what kind of conclusions (universal affirmative, universal negative, particular affirmative, or particular negative) follow from what kind of premises, to enable students of logic to construct valid syllogisms more easily.
The French philosopher Jean Buridan (Joannes Buridanus, c. 1297 – 1358), professor of philosophy in the University of Paris, is credited with devising a set of rules to help slow-witted students in the discovery of syllogistic middle terms, which later became known as the pons asinorum.
In 1480, Petrus Tartaretus applied the Latin expression “pons asinorum” to a diagram illustrating these rules, whose purpose was to help the student of logic find the middle term of a syllogism and disclose its relations to the other terms.
The “asses’ bridge” was usually presented with the predicate, or major term, of the syllogism on the left, and the subject on the right. The three possible relations of the middle term to either the subject or the predicate (consequent, antecedent and extraneous) were represented by six points arranged in two rows of three in the middle of the diagram, between the subject and the predicate. The student was then asked to identify the nineteen valid combinations of the three figures of the syllogism and evaluate the strength of each premise.
Euclid's Fifth Proposition reads:
In isosceles triangles the angles at the base equal one another, and, if the equal straight lines are produced further, then the angles under the base equal one another.
Pappus provided the shortest proof of the first part, that if the triangle is ABC with AB being the same length as AC, then comparing it with the triangle ACB (the mirror image of triangle ABC) will show that two sides and the included angle at A of one are equal to the corresponding parts of the other, so by the fourth proposition (on congruent triangles) the angles at B and C are equal. The difficulty lies in treating one triangle as two, or in making a correspondence, but not the correspondence of identity, between a triangle and itself. Euclid's proof was longer and involved the construction of additional triangles:
In isosceles triangles the angles at the base equal one another, and, if the equal straight lines are produced further, then the angles under the base equal one another. Let ABC be an isosceles triangle having the side AB equal to the side AC, and let the straight lines BD and CE be produced further in a straight line with AB and AC. (Book I.Definition 20; Postulate 2)
I say that the angle ABC equals the angle ACB, and the angle CBD equals the angle BCE. Take an arbitrary point F on BD. Cut off AG from AE the greater equal to AF the less, and join the straight lines FC and GB. (Book I. Proposition 3.; Postulate.1)
Since AF equals AG, and AB equals AC, therefore the two sides FA and AC equal the two sides GA and AB, respectively, and they contain a common angle, the angle FAG.
Therefore the base FC equals the base GB, the triangle AFC equals the triangle AGB, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides, that is, the angle ACF equals the angle ABG, and the angle AFC equals the angle AGB. (Book I.Proposition 4)
Since the whole AF equals the whole AG, and in these AB equals AC, therefore the remainder BF equals the remainder CG. (Common Notion 3)
But FC was also proved equal to GB, therefore the two sides BF and FC equal the two sides CG and GB respectively, and the angle BFC equals the angle CGB, while the base BC is common to them. Therefore the triangle BFC also equals the triangle CGB, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides. Therefore the angle FBC equals the angle GCB, and the angle BCF equals the angle CBG. (Book I. Proposition 4)
Accordingly, since the whole angle ABG was proved equal to the angle ACF, and in these the angle CBG equals the angle BCF, the remaining angle ABC equals the remaining angle ACB, 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.(Common Notion 3)
Therefore in isosceles triangles the angles at the base equal one another, and, if the equal straight lines are produced further, then the angles under the base equal one another.
From D.E. Joyce's presentation of Euclid's ‘‘Elements’’ 
It is the ass's pitfall, not his bridge.
If this be rightly called the “Bridge of Asses,”
He's not the fool who sticks, but he that passes.
All links retrieved June 24, 2016.
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