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Question

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

∴ AB + CD + EF = BC + DE + FA
Hence proved.

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