Euclid, Elements I 5
(Pons asinorum; the bridge of asses)
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translated by Henry Mendell (Cal. State U., L.A.)

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Prop. 5: The angles at the base of isosceles triangles are equal to one another, and when the equal sides are extended the angles under the base will be equal to one another.

(general diagram)

Diagrams for book 1 proposition 5, following the display, construction, and demonstration(diagram 1) Let there be an isosceles triangle, ABG, having side AB equal to side AG, and let straight lines, BD, GE, be extended on a straight line with AB, AG. I say that the angle by ABG is equal to that by AGB and the angle by GBD to that by BGE.


(diagram 2) For let there be taken on BD a point as happens, Z, and from the larger, AE, an equal, AH, to the smaller, AZ, has been taken away, (diagram 3)and let straight-lines ZG, HB be joined.
(diagram 4) And so, since AZ is equal to AH and AB to AG, in fact, two, ZA, AG, are respectively equal to two, HA, AB, And they enclose a common angle, that by ZAH. (diagram 5) Therefore, a base, ZG, is equal to a base, HB, and triangle AZG will be equal to triangle AH, and the remaining angles will be respectively equal to the remaining angles which the equal sides subtend, the angle by AGZ to that by ABH and the angle by AZG to that by AHB. (diagram 6) And since a whole, AZ, is equal to a whole, AH, for which AB is equal to AG, therefore, a remainder, BZ, is equal to GH. (diagram 7) But ZG was also shown equal to HB. In fact, two, BZ, ZG, are respectively equal to two, GH, HB. And an angle, that by BZG, is equal to an angle, that by GHB, and a base of them, BG, is common. (diagram 8) Therefore, triangle BZG is also equal to triangle GHB, and the remaining angles will be respectively equal to the remaining angles which the equal sides subtend. Therefore, the angle by ZBG is equal to that by HGB and the angle by BGZ to that by GBH. (diagram 9) And so, since a whole, the angle by ABH, was shown equal to a whole, that by AGZ, where the angle by GBH is equal to that by BGZ, therefore, a remainder, the angle by ABG, is equal to a remainder, that by AGB. And they are at the base of triangle ABG. But the angle by ZBG was also shown equal to that by HGB. And they are under the base. Therefore, the angles at the base of isosceles triangles are equal to one another, and when the equal sides are extended the angles under the base will be equal to one another, just what it was required to show.<

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