rrrm@cwruecmp.UUCP (R Robertson + R McGuire) (01/31/84)
I don't know how many people have already replied to this 'proof' but
I'm gonna anyhow.
First let me re-run the proof for those of you that missed it...
1 Construct triangle ABC
2 Bisect angle BAC
3 Perpendicularly bisect BC; D is at the midpoint of BC.
4 E is where BC's bisector and angle A' bisector cross.
(if the two bisectors are coincident then E is any point on the line.)
5 draw BE & CE
6 F is the projection of E onto AB
7 G is the projection of E onto AC
8 BD=CD (step 3); BDE=CDE=90 (steps 3&4); ED=ED (reflexive);
therefore triangle BDE=triangle CDE.
9 FAE=GAE (step 2); AFE=AGE=90 (steps 6&7); AE=AE (reflexive);
therefore triangle AFE=triangle AGE.
10 EFB=EGC=90 (steps 6&7); FE=GE (step 9); BE=CE (step 8);
therefore triangle EFB=triangle EGC
11 AF=AG (step 9); FB=GC (step 10); AF+FB=AB; AG+GC=AC
therefore AB=AC
therefore triangle ABC is isosceles.
12 Since it doesn't matter how you label the triangle in step 1, you can
prove that ANY two of the sides are the same length; therefore ALL the
sides are the same length and the triangle is equilateral.
Q.E.D.
Wasn't that fun. Try a few to see if you can tell what's wrong before
you read the rest of this article.
Tried one? Good, now I'll tell you the problem (just in case you
didn't figure it out for yourself.
The problem is in step 11. If AF+FB=AB and AG+GC=AC then the triangle
is indeed isosceles with A as the vertex. More often than not, however,
if AF+FB=AB then AC+CG=AG, or if AG+GC=AC then AB+BF=AF. (i.e. F and G
aren't neccisarily between A&B and A&C, respectively.) Well, that's
it.
Ryan McGuire
(the rm of rrrm)