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.

Chuck