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

**Ans: **Let’s first draw a diagram for the given question:

Let ABCDEF be the Heaxagon which is circumscribed on a circle. This circle has Center O and radius R.

Now let P, Q, R, S, T and U be the points on which hexagon touches the circle.

**Step 1:** Let’s start with point A.

AP and AQ are the tangents drawn from an external point A on same circle, hence, AP = AQ

Similarly, BP and BU are the tangents drawn from an external point B on same circle, hence, BP = BU

Similarly, CU = CT, DT = DS, ES = ER and FR = FQ

**Step 2:** Now, AB + CD + EF = (AP + BP) + (CT + DT) + (ER + FR)

= (AQ + BU) + (CU + DS) + (ES + FQ) (by substituting values from step 1 identified above)

= (BU + CU) + (DS + ES) + (FQ + AQ)

= BC + DE + FA

**Hence Proved !**

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