Re: geometry problem (triangle)

In article <dbbd541oni54g6od45k7mrlbpgk947n2pn@xxxxxxx>,
Angus Rodgers <twirlip@xxxxxxxxxxx> wrote:

....
Lemma. If P is inside the triangle ABC, then PB + PC < AB + AC.

Proof. Draw B'C' through P parallel to BC, with B' on AB and C' on
AC. Then, by the Triangle Inequality:

PB + PC < (PB' + B'B) + (PC' + C'C)

= B'B + (B'P + PC') + C'C

= B'B + B'C' + C'C

< B'B + (B'A + AC') + C'C

= (AB' + B'B) + (AC' + C'C)

= AB + AC

Q.E.D.
....


Having recently returned home and found this thread, I can't
resist adding a belated historical comment. The lemma above is Euclid
I.21, and Angus may agree that Euclid's proof is slightly neater.

Ken Pledger.
.