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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