If a hexagon ABCDEF circumscribe a circle. Prove that
AB+CD+EF = BC+DE+FA
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Solution
Given a hexagon ABCDEF circumscribe a circle.
To prove AB+CD+EF = BC+DE+FA
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