Question

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

Answer 1

Step 1: Understand the given hexagon:
We are given a hexagon $ABCDEF$ that circumscribes a circle. This implies the circle is tangent to each side of the hexagon. A property of circles is that tangents drawn from an external point to a circle are equal in length.
The objective is to prove:
\[AB + CD + EF = BC + DE + FA\]

Step 2: Use the properties of tangents from external points:
Let the points of tangency on sides $AB, BC, CD, DE, EF, FA$ be $P, Q, R, S, T, U$ respectively. Tangents from an external point to a circle are equal. Therefore:
- Tangents from $A$ are $AP = AS$.
- Tangents from $B$ are $BP = BQ$.
- Similarly for other vertices.

Step 3: Assign variables to the tangent segments:
Let the lengths of the tangent segments from each vertex be:
- $AP = AS = x$
- $BP = BQ = y$
- $CQ = CR = z$
- $DR = DS = w$
- $ER = ET = p$
- $FU = FA = q$

Step 4: Express the lengths of the sides in terms of the tangent segments:
The lengths of the hexagon's sides can be expressed as the sum of two tangent segments:
- $AB = AP + BP = x + y$
- $BC = BQ + CQ = y + z$
- $CD = CR + DR = z + w$
- $DE = DS + ER = w + p$
- $EF = ET + FR = p + q$
- $FA = FU + AS = q + x$

Step 5: Add the sides of the hexagon:
Sum of alternating sides:
\[AB + CD + EF = (x + y) + (z + w) + (p + q) = x + y + z + w + p + q\]Sum of the other alternating sides:
\[BC + DE + FA = (y + z) + (w + p) + (q + x) = x + y + z + w + p + q\]Comparing the two sums:
\[AB + CD + EF = BC + DE + FA\]

Conclusion:
The equality has been proven:
\[AB + CD + EF = BC + DE + FA\]This concludes the proof.