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

**Solution:**

Given, a __hexagon__ ABCDEF circumscribe a circle

We have to prove that AB + CD + EF = BC + DE + FA

We know that the __tangents__ to a circle through an external point are equal.

From the figure,

AR and AM are the tangent to the circle through an point A

So, AM = AR -------------------- (1)

Similarly,

The tangents through point B

BM = BN ------------------------ (2)

The tangents through point C

CO = CN ------------------------ (3)

The tangents through point D

DO = DP ------------------------ (4)

The tangents through point E

EQ = EP ------------------------ (5)

The tangents through point F

FQ = FR ------------------------ (6)

Adding (1) to (6),

AM + BM + CO + DO + EQ + FQ = AR + BN + CN + DP + EP + FR

On rearranging,

(AM + BM) + (CO + DO) + (EQ + FQ) = (AR + FR) + (BN + CN) + (EP + DP)

From the figure,

AM + BM = AB

CO + DO = CD

EQ + FQ = EF

AR + FR = AF

BN + CN = BC

EP + DP = ED

So, AB + CD + EF = AF + BC + ED

Therefore, AB + CD + EF = BC + DE + FA

**✦ Try This: **Suppose ABCDEF is a hexagon such that AB = BC = CD = 1 and DE = EF = FA = 2. If the vertices A, B, C, D, E, F are concyclic, the radius of the circle passing through them is

**☛ Also Check:** NCERT Solutions for Class 10 Maths Chapter 10

**NCERT Exemplar Class 10 Maths Exercise 9.4 Problem 1**

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

**Summary:**

Hexagon is a two-dimensional shape with six sides, six vertices, and six edges. If a hexagon ABCDEF circumscribes a circle, it is proven that AB + CD + EF = BC + DE + FA

**☛ Related Questions:**

- Let s denote the semi-perimeter of a triangle ABC in which BC = a, CA = b, AB = c. If a circle touch . . . .
- From an external point P, two tangents, PA and PB are drawn to a circle with centre O. At one point . . . .
- If AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in . . . .

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