> Books I have referenced mention various methods of > proving triangle similarity. One of these is that, if > two sides of a triangle are proportional to the > corresponding sides of another triangle, and the > included angles in each triange are equal, then the > triangles are similar. It seems to me that if two > sides of a triangle are proportional to corresponding > sides of a second triangle, and a not-included angle > is equal to the the corresponding not-included angle > of the other triangle, then the triangles are similar > also. But an examination of a few text books do not > mention this as a theorem. In fact, one text has this > as a problem, and says the triangles cannot be > deemed similar in the second case. Any opinions out > there?
The reason it is not mentioned is that it is not true in general. Consider triangle ABC with <A = 30, AB = 10, BC = 6. Now consider triangle DEF with <D = 30, DE = 5, and EF = 3. You can draw triangle DEF with <F acute or with <F obtuse (actually you can do the same thing with <C). The situation is kind of like the ambiguity that arises with SSA as a supposed method of congruence. The shape of the triangle is not uniquely determined.