One minor point to add to modellingman's answer: he starts it with "The triangle has sides a, b and c in decreasing order of length." but it isn't immediately obvious where he uses that order. The following proof (which I must have first been told around 45 years ago, and it was clearly nowhere near new then) might provide some insight about that.
Consider a triangle ABC. Draw the bisector of the angle at A and the perpendicular bisector of the side BC. Assuming they're not parallel, they meet at a single point - call it X. In diagrammatic form (and please excuse the diagram not being to scale - ASCII diagrams are a somewhat limiting medium, and a proper embedded image is rather more trouble than it's worth, as well as not working well with spoiler protection if anyone wants to use it in a solution):
. \
\
_A__
_- \ --__
_- \ --__
_- \ --__
_- \ --__
_- \ --__
_- \ --__
_- \ --__
_- \ --__
_- \ --__
_- \ --__
_- \ --__
_- \ | --__
_- \ | --__
B------------------------------------------\-----D-------------------------------------C
\ |
\ |
\ |
\ |
\|
X
|\
| \
in which angle BAX = angle CAX (since the line AX is the bisector of angle BAC), D is the midpoint of line BC and angles ADX and CDX are both right angles (since the line DX is the perpendicular bisector of line BC).
Now drop perpendiculars from X to the sides AB and AC and label the points where they intersect those sides as Y and Z respectively:
. \
\
_A__
_- \ --__
_- \ --__
-_ _- \ --__
-_ _- \ --__
Y- \ --__
_- -_ \ --__ /
_- -_ \ --__ /
_- -_ \ -Z__
_- -_ \ / --__
_- -_ \ / --__
_- -_ \ / --__
_- -_ \ | / --__
_- -_ \ | / --__
B------------------------------------_-----\-----D-----/--------------------------------C
-_ \ | /
-_ \ | /
-_ \ | /
-_ \ | /
-_\|/
X_
/|\-_
/ | \ -
in which angles AYX, BYX, AZX and CZX are all right angles, because the lines XY and XZ were constructed to be perpendicular to the lines AB and AC respectively.
Now we have angle YAX = angle ZAX (by the construction of X in the first step above), angle AYX = angle AZX (by the construction of Y and Z in the second step) and distance AX in the AXY triangle = distance AX in the AXZ triangle (obvious), from which it follows that triangles AYX and AZX are similar, and thus that distance AY = distance AZ and distance XY = distance XZ. (Yes, I know that those conclusions look blatantly wrong in the ASCII diagram, but as I said, please excuse that - it's very hard to draw this ASCII diagram to be even as roughly to scale as the pencil-and-paper sketch I was shown ~45 years ago...)
But now we have distance XB = distance XC (because X is on the perpendicular bisector of BC), distance XY = distance XZ (just proved) and angle XYB = angle XZC (since they were both constructed to be right angles in the second step above). So triangles XYB and XZC are also similar to each other, from which it follows that distance YB = distance ZC.
But now we have distance AB = distance AY + distance YB = distance AZ + distance ZC = distance AC.
There is the remaining possibility that the bisector of the angle at A and the perpendicular bisector of the side BC are parallel and so don't meet at a point (or meet at every point along them), so that the point X hasn't actually been defined. But in that case, the bisector of angle A meets the side BC perpendicularly at a point E, say, and we have the following diagram (still not to scale):
. |
|
_A__
_- | --__
_- | --__
_- | --__
_- | --__
_- | --__
_- | --__
_- | --__
_- | --__
_- | --__
_- | --__
_- | --__
_- | --__
_- | --__
B----------------------------E---------------------------------------------------------C
|
|
But we now have angle BEA = angle CEA (since AE is perpendicular to BC), angle BAE = angle CAE (since AE is the bisector of BC) and distance AE in triangle BAE = distance AE in triangle CAE. It follows that triangles BAE and CAE are similar, so that distance AB = distance AC. (And also that distance BE = distance CE, so in fact this case is when the bisector of the angle at A
is the perpendicular bisector of BC and not merely parallel to it. So I could actually have drawn the diagram to be more to scale - but doing so would have surreptitiously assumed the answer, so I reckoned it wasn't a good idea.)
So in either case, I've proved that distance AB = distance AC - i.e. that the arbitrary triangle ABC I started with must have two equal sides. O more briefly, that all triangles are isosceles triangles.
Or have I? If not, where's the flaw in the proof?
Gengulphus