From: Bill Dubuque
Newsgroups: sci.math.research
Subject: [conway@math.Princeton.EDU: Re: Geometrical "proof"]
Date: Tue, 5 Aug 1997 01:34:33 -0400
Somehow Conway's addendum to the message below made it to
sci.math.research (from math-fun) but the message below did not.
-Bill Dubuque
------- Start of forwarded message -------
Date: Sun, 3 Aug 1997 11:40:05 -0400 (EDT)
From: John Conway
Subject: Re: Geometrical "proof"
To: Bill Dubuque
Cc: tors@math.uio.no, math-fun@cs.arizona.edu
On 2 Aug 1997, Bill Dubuque wrote:
> tors@math.uio.no (Thor Sandmel) writes:
> |
> | I remember having seen a "proof" that all triangles are isosceles, but now
> | I cannot find it. Anybody got a reference? Other examples of geometrical
> | "proofs" made incorrect by improper reference to a figure are also welcome
> | (but they must be short and simple). Grateful for any hint. Please respond
> | by e-mail.
This is old, and is often supposed to have come from Pappus' lost
book of porisms, although the earliest place I've seen it is in
something of Lewis Carrol's. Here it is. I have to draw the triangle
as equilateral, but that's no loss since, as the theorem shows, every
triangle actually is!
B
/|\
c/ | \a
/ O \
/___|___\
A b C
O is where the angle-bisector BO of B meets the perpendicular
bisector bO of the edge AB. Then a and c are the feet of the
perpendiculars from O to the sides BC and AB. Then we can see the
congruences:
OcB == OaB (right angle, hyp and another side)
OCb == OAb (two sides and included (right) angle)
AOc == COa (right angle, hyp and another side)
which show that indeed the triangle is isosceles (and, by repeating
the argument, equilateral).
John Conway
------- End of forwarded message -------
==============================================================================
Newsgroups: sci.math
From: eclrh@sun.leeds.ac.uk (Robert Hill)
Subject: Re: Amusing 'Proofs'
Date: Fri, 13 Mar 1998 17:25:43 +0000 (GMT)
In article , Michael Anttila writes:
> One of my favourite proofs is the proof that all triangles are isosceles
> (i.e. have two equal sides for the layman). It's hard to do in a message
> without giving away the secret, but when done on a blackboard it can be
> pretty convincing (had me stumped for a while). Here's basically how it
> goes:
>
> - Draw a scalene triangle like so (forgive my ascii art - and read with a
> fixed-width font): Make sure you do all the drawing freehand. If you
> measure stuff out or use a compass, you'll screw up the "proof".
> A
> _-/
> _-~ /
> _-~ /
> _-~ /
> _-~ /
> B ~~~~~~~~~~~ C
>
> - Now, construct a perpendicular bisector of side BC, and a bisector of
> angle A, like so:
> A
> _-/
> _-~//
> _-~ /~/ OK, you really have to use your imagination here.
> _-|X/~~ /
> _-~ |~ /
> B ~~~~~~~~~~~ C
>
> Anyway, they meet at some point in the middle of the triangle. Call it X.
>
> - Now drop down perpendiculars from X to sides AB and AC. From here on in
> it should be a simple matter using similar triangles and basic angle
> identities to show that side AB = side AC. Therefore all triangles are
> isosceles.
>
> The reason why this doesn't work should be pretty obvious from my
> description, but if you can't figure it out, the solution is below
> (separated by spoiler space)...
> Well, it's because the perpendicular bisector of BC and the bisector of
> angle A don't meet inside the triangle - except in one case: When the
> triangle is isosceles. However, if you're drawing it freehand, you can
> use the excuse that your artistic skills are bad and the argument becomes
> pretty convincing.
The "proof" can easily be extended so that it appears to include the case
where the point X is outside the triangle. It is often presented as saying:
"If X is inside, use this argument; if outside, use that one. Both ways
the triangle is isosceles". But when it is so modified, it depends on
another false assumption: that the feet of the perpendiculars from X
to AB, AC are both outside the triangle.
--
Robert Hill
University Computing Service, Leeds University, England
"Though all my wares be trash, the heart is true."
- John Dowland, Fine Knacks for Ladies (1600)