For, let BB and GE meet at; then if the angle OBC be not equal to the angle OCB one of them must be greater than the other; let the angle OBC be the greater. Then, because CB and BD are equal to BC and CE, each to each; but the angle CBD is greater than the angle BCE; therefore CD is greater than BE (I. 24).
On the other side of the base BC make the triangle BCF equal to the triangle CBE, so that BF may be equal to CE, and CF equal to BE (I. 22); and join DF.
Then because BF is equal to BD the angle BED is equal to the angle BDF. And. the angle OCD is, by hypothesis, less than the angle OBE; and the angle COD is equal to the angle BOE; therefore the angle ODC is greater than the angle OEB (I, 32), and therefore the angle ODC is greater than the angle BFC.
Hence, by taking away the equal angles BDF and BFD, the angle FDC is greater than the angle DFC; and therefore CF is greater than CD (I. 1 9); therefore BE is greater than CD.
But it was shewn that CD is greater than BE which is absurd.
Therefore the angles OBC and OCB are not unequal, that is, they are equal; and therefore the angle ABC is equal to the angle ACB.
[For the history of this theorem see Lady's and Gentleman's Diary for 1869, page 88.]
