Question

if hexagon ABCDEF circumscribe a circle prove that AB+CD+EF=BC+DE+FA

Answer 1

Solution:-

Given : Here ABCDEF is a hexagon which circumscribes a circle.
Since the tangents drawn from a point to the circle are equal in length.
Let the inside points where the circle touches the hexagon be 'M', 'N', 'O', 'P', 'Q' and 'R' respectively.
Therefore,
AM = AR ...(1)
BM = BN ...(2)
CN = CO ...(3)
DO = DP ...(4)
EP = EQ ...(5)
FQ = FR ...(6)
Now, from these equations, we have
AM + BM = AR + BN ⇒ AB = AR + BN
CO + DO = CN + DP ⇒ CD = CN + DP
EQ + FQ = EP + FR ⇒ EF = EP + FR
Now, on adding the above three, we get
AB + CD + EF = AR + (BN + CN) + (DP + EP) + FR = BC + DE + FA
⇒ AB + CD + EF = BC + DE + FA
Hence proved.

Answer 2

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