Re: proof by contradiction in Euclid's elements
- From: William Elliot <marsh@xxxxxxxxxxxxxxxxxx>
- Date: Tue, 26 Apr 2005 20:01:57 -0700
On Tue, 26 Apr 2005, ken quirici wrote:
> PROP: If in a triangle two angles equal one another, then the sides
> opposite the equal angles also equal one another.
>
> Let ABC be a triangle having the angle ABC equal to the angle ACB.
> I say that the side AB also equals the side AC.
>
Let AD bisect angle BAC with D on BC. Show ADB and ADC congruent.
> If AB does not equal AC, then one of them is greater.
>
Bah.
> Let AB be greater. Cut off DB from AB the greater equal to AC the
> less, and join DC.
>
> Since DB equals AC, and BC is common, therefore the two sides DB and
> BC equal the two sides AC and CB respectively, and the angle DBC
> equals the angle ACB. Therefore the base DC equals the base AB, and
> the triangle DBC equals the triangle ACB, the less equals the greater,
> which is absurd. Therefore AB is not unequal to AC, it therefore
> equals it.
>
> Therefore if in a triangle two angles equal one another, then the
> sides opposite the equal angles also equal one another.
.
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