Question

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

Answer 1

Given: ATQ, A Hexagon ABCDEF is circumscribe a circle.

To prove: AB + CD + EF = BC + DE + FA

Proof: Tangents drawn from an external point to a circle are equal.

AM = RA - (I) {Tangents from point A}

BM = BN - (II) {Tangents from point B}

CO = NC - (III) {Tangents from point C}

OD = DP - (IV) {Tangents from point D}

EQ = PE - (V) {Tangents from point E}

QF = FR - (VI) {Tangents from point F}

Now, all equations (I) to (VI),

Hence proved.