Question

if an hexagon abcdef circumscribe a circle prove that ab+cd+ef=bc+de+fa

Answer 1

a hexagon ABCDEF circumscribes a circle, prove that AB+CD+EF=BC+DE+FA. 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.

Answer 2

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