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

To prove that in a hexagon ABCDEF that circumscribes a circle, the relationship AB + CD + EF = BC + DE + FA holds, we can use the properties of tangents from a point to a circle.

Understanding Tangents

When a hexagon circumscribes a circle, each side of the hexagon is tangent to the circle. This means that the lengths of the tangents drawn from each vertex to the points of tangency are equal.

Assigning Tangent Lengths

Let’s denote the points where the circle touches the sides of the hexagon as follows:

• Point P touches side AB
• Point Q touches side BC
• Point R touches side CD
• Point S touches side DE
• Point T touches side EF
• Point U touches side FA

Now, we can assign the following tangent lengths:

• AP = AU = x
• BP = BQ = y
• CQ = CR = z
• DR = DS = w
• ET = ES = v
• FT = FU = u

Expressing Side Lengths

Using these tangent lengths, we can express the lengths of the sides of the hexagon:

• AB = AP + BP = x + y
• BC = BQ + CQ = y + z
• CD = CR + DR = z + w
• DE = DS + ET = w + v
• EF = ET + FT = v + u
• FA = FU + AU = u + x

Summing the Sides

Now, let’s sum the sides according to the required relationship:

• AB + CD + EF = (x + y) + (z + w) + (v + u)
• BC + DE + FA = (y + z) + (w + v) + (u + x)

When we simplify both sides, we find:

• AB + CD + EF = x + y + z + w + v + u
• BC + DE + FA = x + y + z + w + v + u

Conclusion

Since both expressions are equal, we have shown that AB + CD + EF = BC + DE + FA. This completes the proof for the hexagon that circumscribes a circle.