Commentary on Proposition XIII.7

Back to Proposition XIII.7.

This proposition is required in XIII. 17. The steps of the proof may be shown thus.

  1. Suppose that the angles at A, B, C are all equal. Then the isosceles triangles BAE, ABC are equal in all respects; thus BE $= $ AC, $\angle$ BCA $= \angle$ BEA, $\angle$ CAB $= \angle$ EBA. By the last equality, FA $= $ FB, so that, since BE $= $ AC, FC $= $ FE. The $\Delta\text{s}$ FED, FCD are now equal in all respects, [I. 8, 4] and $\angle$ FCD $= \angle$ FED. But $\angle$ ACB $= \angle$ AEB, from above, whence, by addition, $\angle$ BCD $= \angle$ AED. Similarly it may be proved that $\angle$ CDE is also equal to any one of the angles at A, B, C.

  2. Suppose the angles at A, C, D to be equal. Then the isosceles triangles ABE, CBD are equal in all respects, and hence BE $= $ BD (so that $\angle$ BDE $= \angle$ BED), and $\angle$ CDB $= \angle$ AEB. By addition of the equal angles, $\angle$ CDE $= \angle$ DEA. Similarly it may be proved that $\angle$ ABC is also equal to each of the angles at A, C, D.