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 means that the circle is tangent to each of the sides of the hexagon, and the tangents from any external point to a circle are equal in length.
We are asked to prove that:
\[ AB + CD + EF = BC + DE + FA \] This is the relationship we need to prove.

Step 2: Use the properties of tangents from external points:
Let the points of tangency of the circle with the sides of the hexagon be $P$, $Q$, $R$, $S$, $T$, and $U$, where each letter corresponds to the side of the hexagon that is tangent to the circle. The tangents drawn from an external point to a circle are equal in length.
Thus, the following relationships hold:
- The length of the tangent from $A$ to the circle is equal to the length of the tangent from $B$ to the circle, so let the tangents from $A$ and $B$ be $AP = AS$ and $BP = BQ$ respectively.
- Similarly, the lengths of the tangents from the other points are also equal.

Step 3: Assign variables to the tangent segments:
Let the tangents from the points of the hexagon to the circle be:
- $AP = AS = x$
- $BP = BQ = y$
- $CQ = CR = z$
- $DR = DS = w$
- $ER = ET = p$
- $FR = FU = q$

Step 4: Express the lengths of the sides in terms of the tangent segments:
Using the fact that the tangents from a point to the circle are equal, we can express the sides of the hexagon as:
- $AB = AP + BP = x + y$
- $BC = BQ + CQ = y + z$
- $CD = CR + DR = z + w$
- $DE = DS + ER = w + p$
- $EF = ET + FR = p + q$
- $FA = FU + AS = q + x$

Step 5: Add the sides of the hexagon:
Now, add the sides $AB$, $CD$, and $EF$:
\[ AB + CD + EF = (x + y) + (z + w) + (p + q) = x + y + z + w + p + q \] Similarly, add the sides $BC$, $DE$, and $FA$:
\[ BC + DE + FA = (y + z) + (w + p) + (q + x) = x + y + z + w + p + q \] Thus, we have:
\[ AB + CD + EF = BC + DE + FA \]

Conclusion:
We have proved that:
\[ AB + CD + EF = BC + DE + FA \] This completes the proof.