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 from an external point:
$AP = AU$
$BP = BQ$
$CQ = CR$
$DR = DS$
$ES = ET$
$FT = FU$

Now, consider the sum $AB + CD + EF$:
$AB = AP + PB$
$CD = CR + RD$
$EF = ET + TF$

So,
$AB + CD + EF$
$= (AP + PB) + (CR + RD) + (ET + TF)$

Substitute equal tangent segments:
$= (AU + BQ) + (CQ + DS) + (ES + FU)$

Rearranging terms:
$= (BQ + CQ) + (DS + ES) + (FU + AU)$

$= BC + DE + FA$

Therefore,
$AB + CD + EF = BC + DE + FA$
Step 4: Final Answer:
Hence proved.