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Proclus' variation of Euclid's proof proceeds as follows: Let ABC be an isosceles triangle with AB and AC being the equal sides. Pick an arbitrary point D on side AB and construct E on AC so that AD=AE. Draw the lines BE, DC and DE. Consider the triangles BAE and CAD; BA=CA, AE=AD, and angle A is equal to itself, so by side-angle-side, the triangles are congruent and corresponding sides and angles are equal. Therefore angle ABE = angle ACD, angle ADC = angle AEB, and BE=CD. Since AB=AC and AD=AE, BD=CE by subtraction of equal parts. Now consider the triangles DBE and ECD; BD=CE, BE=CD, and angle DBE = angle ECD have just been shown, so applying side-angle-side again, the triangles are congruent. Therefore angle BDE = angle CED and angle BED = angle CDE. Since angle BDE = angle CED and angle CDE = angle BED, angle BDC = angle CEB by subtraction of equal parts. Consider a third pair of triangles, BDC and CEB; DB=EC, DC=EB, and angle BDC = angle CEB, so applying side-angle-side a third time, the triangles are congruent. In particular, angle CBD = BCE, which was to be proved.

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