Question

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

Hint:

For any polygon circumscribing a circle, alternating side sums are often equal if the polygon has certain symmetries or an even number of sides.

Answer 1

Step 1: Understanding the Concept:
The lengths of tangents drawn from an external point to a circle are equal.
Step 2: Key Formula or Approach:
Let the circle touch the sides $AB, BC, CD, DE, EF, FA$ at points $P, Q, R, S, T, U$ respectively.
Step 3: Detailed Explanation:
By the property of tangents:
$AP = AU$, $BP = BQ$, $CQ = CR$, $DR = DS$, $ES = ET$, $FT = FU$.
Now, let's sum the segments for $AB + CD + EF$:
$AB + CD + EF = (AP + PB) + (CR + RD) + (ET + TF)$
Substitute the equal tangent segments:
$= (AU + BQ) + (CQ + DS) + (ES + FU)$
Rearranging the terms:
$= (BQ + CQ) + (DS + ES) + (FU + AU)$
$= BC + DE + FA$.
Thus, $AB + CD + EF = BC + DE + FA$.
Step 4: Final Answer:
Hence proved.