If a hexagon $ABCDEF$ circumscribes a circle,prove that $AB + CD + EF = BC + DE + FA$.

(N/A) Let the points of contact of the circle with the sides $AB, BC, CD, DE, EF,$ and $FA$ be $P, Q, R, S, T,$ and $U$ respectively.
Since the lengths of tangents drawn from an external point to a circle are equal,we have:
$AP = AU$
$BP = BQ$
$CQ = CR$
$DR = DS$
$ES = ET$
$FT = FU$
Now,consider the sum of alternate sides:
$AB + CD + EF = (AP + PB) + (CR + RD) + (ET + TF)$
Substituting the tangent equalities:
$AB + CD + EF = (AU + BQ) + (CQ + DS) + (ES + FU)$
Rearranging the terms:
$AB + CD + EF = (BQ + CQ) + (DS + ES) + (FU + AU)$
$AB + CD + EF = BC + DE + FA$
Hence,it is proved.