Date: Aug 20, 2012 7:14 PM
Author: Chuck
Subject: Re: Triangle similarity

> 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.