the angle ACB is equal to the angle ABC, by hypothesis. Or I. 6 might be placed after I. 16 and demonstrated thus. Bisect the angle BAC by a straight line meeting the base at D. Then the triangles ABD and ACD are equal in all respects, by I. 16.
I. 7 is only required in order to lead to I. 8. The two might be superseded by another demonstration of I. 8, which has been recommended by many writers.
Let ABC, DEF be two triangles, having the sides AB, AC equal to the sides DB, DF, each to each, and the base BC equal to the base EF: the angle BAC shall be equal to the angle EDF.
For, let the triangle DEF be applied to the triangle ABC, so that the bases may coincide, the equal sides be conterminous, and the vertices fall on opposite sides of the base. Let GBC represent the triangle DEF thus applied, so that G corresponds to D. Join AG. Since, by hypothesis, BA is equal to BG, the angle BAG is equal to the angle BGA, hy I. 5. In the same manner the angle CAG is equal to the angle CGA. Therefore the whole angle BAC is equal to the whole angle BGC, that is, to the angle EDF.
There are two other cases; for the straight line AG may pass through B or C, or it may fall outside BC: "these cases may be treated in the same manner as that which we have considered.
I. 8. It may be observed that the two triangles in I. 8 are equal in all respects; Euclid however does not assert more than the equality of the angles opposite to the bases, and when he requires more than this result he obtains it by using I. 4.
I. 9. Here the equilateral triangle DEF is to be described on the side remote from A, because if it were described on the same, side, its vertex, F, might coincide with A, and then the construction would fail.
