If a hexagon abcdef circumscribed a circle proved that ab+cd+ef=bc+de+

1.	If a hexagon abcdef circumscribed a circle proved that ab+cd+ef=bc+de+fa
Answer» Sol: Let ABCDEF be a hexagon.It is circumscribe a circle. But length of the tangents drawn from a point to the circle are equal. AM = AR -----(i) BM = BN ---------(ii) CN = CO ----------(iii) DO = DP ----------(iv) EP = EQ ----------(v) FQ = FR --------- (vi) adding (i) and (ii) we get AM + BM = AR + BN ⇒ AB = AR + BN adding (iii) and (iv) we get CO + DO = CN + DP ⇒ CD = CN + DP adding (v) and (vi) we get EQ + FQ = EP + FR ⇒ EF = EP + FR Adding all these we obtain AB + CD + EF = AR + ( BN + CN ) + (DP + EP) + FR = BC + DE + FA ∴ AB + CD + EF = BC + DE + FA

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