[net.math] Proof that all triangles are equilateral

wolit@rabbit.UUCP (01/23/84)

Theorem:  All triangles are equilateral.

Construction (refer to diagram below):

1.  Construct triangle ABC.
2.  Construct the bisector of angle BAC ("." line).
3.  Label the bisector of side BC as "D" and construct a perpendicular
    to BC at D ("|" line).
4.  Label the intersection of these two lines (the bisector of angle BAC
    and the perpendicular bisector of side BC) as "E".
    [Note:  in a *REAL* isosceles triangle, these lines are coincident.
    In that case, choose any point along the line AD, and call it "E".]
5.  Construct line segments BE and CE ("," lines).
6.  Construct a perpendicular to AB through E; label it F ("-" line).
7.  Construct a perpendicular to AC through E; label it G ("-" line).

                                   A
                                  /.\
                                 / . \
                                /  .  \
                               /   .   \
                              /    .    \
                             /     .     \
                            /      .      \
                           /       .       \
                          /        .        \
                         /         .         \
                        F-         .         -G
                       /    -      .      -    \
                      /        -   .   -        \
                     /             E             \
                    /           ,  |  ,           \
                   /         ,     |     ,         \
                  /       ,        |        ,       \
                 /     ,           |           ,     \
                /   ,              |              ,   \
               / ,                 |                 , \
              B____________________D____________________C

Proof:

8.  Consider right triangles BDE and CDE.  Side BD = DC, and 
    angle BDE = CDE = 90 degrees (from 3), and, of course, side
    ED = ED.  Therefore, triangles BDE and CDE are congruent, and
    hypotenuse BE = CE.
9.  Consider right triangles AFE and AGE.  Angle FAE = GAE (from 2).
    Angle AFE = AGE = 90 degrees (from 6 and 7).  Hypotenuse AE = AE.
    Therefore, the triangles AFE and AGE are congruent, and side
    AF = AG, and side FE = GE.
10. Consider right triangles EFB and EGC.  Angle EFB = EGC = 90
    degrees (from 6 and 7).  Side FE = GE (from 9).  Hypotenuse BE = CE
    (from 8).  Therefore, the triangles EFB and EGC are congruent, and
    side FB = GC.
11. Consider the original triangle ABC.  Side AB = AF + FB, and side
    AC = AG + GC.  But AF = AG (from 9) and FB = GC (from 10).
    Therefore, AB = AC, i.e., the triangle is isosceles.
12. But we placed no special conditions on our original choice of A as
    the vertex of this triangle, i.e., we could have proved any two
    sides of the original triangle to be equal.  Therefore, all three
    sides are equal, i.e., the triangle is equilateral.  Since we
    placed no special conditions on the construction of the triangle,
    *ALL* triangle are therefore equilateral.

    Q.E.D.

	Jan Wolitzky, AT&T Bell Laboratories, Murray Hill, NJ

rbd@ihuxr.UUCP (01/24/84)

 The only flaw in this proof is of course that point E need not be inside
the triangle.
                                                  Bob Dianda