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
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