If a hexagon ABCDEF circumscribes a circle, prove that
\(AB + CD + EF = BC + DE + FA\).
Relation is proved: \(AB + CD + EF = BC + DE + FA\).
Step 1: Recall the property of tangents.
If two tangents are drawn to a circle from the same external point, their lengths are equal.
For example, from point A the tangents are to sides AB and AF. So, we can say:
Let \(AP = AF'\) and \(AQ = AB'\). Actually, just note: AP = AF and AQ = AB.
Step 2: Apply this property at each vertex of the hexagon.
Step 3: Write the sides of the hexagon in terms of these tangent lengths.
Step 4: Now, add the sides according to the relation we need to prove.
Left side: \(AB + CD + EF\)
= (x + y) + (z + p) + (q + r)
= x + y + z + p + q + r
Right side: \(BC + DE + FA\)
= (y + z) + (p + q) + (r + x)
= x + y + z + p + q + r
Step 5: Compare both sides.
They are equal: \(AB + CD + EF = BC + DE + FA\).
Final Result: The relation is proved.