[TYPES] A geometry problem for proof systems
Andrew Myers
andru at cs.cornell.edu
Thu Sep 22 08:35:14 EDT 2005
On Thu, Sep 22, 2005 at 06:53:26AM -0400, Peter Freyd wrote:
> These days, of course, one just asks Google. The query:
>
> angle bisector isosceles
>
> leads one to
>
> www.mathpages.com/home/kmath433.htm
>
> It has the ugly proof.
>
> Peter
Here is a more geometric proof:
Theorem:
Given triangle ABC, with angle bisectors BD and EC, angles b = DBC
and c = ECB, show that BD = EC --> ABC is isoceles.
Lemma 1:
Given two triangles ABC and DEF,
If AB = DE & BC = EF, then angle ABC > angle DEF <--> AC > DF
Proof of Lemma 1:
From the law of cosines, we have
AC^2 = AB^2 + BC^2 - 2(AB)(BC)cos(ABC)
DF^2 = DE^2 + EF^2 - 2(DE)(EF)cos(DEF)
= AB^2 + BC^2 - 2(AB)(BC)cos(DEF)
Now, ABC > DEF <--> cos(ABC) < cos(DEF) <--> AC^2 > DF^2 <--> AC > DF.
Proof of Theorem:
Assume that b > c.
1. Apply Lemma 1 to trangles DBC and ECB, showing that DC > EB.
2. Construct point F such that BEFD is a parallelogram, EF = BD &
EB = DF.
3. Since EF = BD = EC, triangle EFC is isoceles and EFC = ECF.
4. Therefore EFD + DFC = DCE + DCF.
5. EFD = EBD = b > c = DCE, so EFD > DCE and DFC < DCF.
6. By Lemma 1 using DFC and DCF, we have DC < DF = BE, which contradicts
step 1.
Therefore b <= c. Symmetrically, c <= b, so b = c, ABC = ACB,
and ABC is isoceles.
--
Andrew Myers
Associate Professor
Computer Science Department
Cornell University
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