Prove that the parallelogram circumscribing a circle is a rhombus
ABCD is a parallelogram. Therefore, opposite sides are equal.
AB = CD
BC = AD
BP = BQ (Tangents from point B)…… (1)
CR = CQ (Tangents from point C)…… (2)
DR = DS (Tangents from point D)…… (3)
AP = AS (Tangents from point A)……. (4)
Adding (1) + (2) + (3) + (4)
BP + CR + DR + AP = BQ + CQ + DS + AS
BP + AP + CR + DR = BQ + CQ + DS + AS
AB + CD = BC + AD
Substitute CD = AB and AD = BC since ABCD is a parallelogram, then
AB + AB = BC + BC
2AB = 2BC
AB = BC
∴ AB = BC = CD = DA
This implies that all the four sides are equal.
Therefore, the parallelogram circumscribing a circle is a rhombus.
Prove that the parallelogram circumscribing a circle is a rhombus.
NCERT Solutions Class 10 Maths Chapter 10 Exercise 10.2 Question 11
It has been proved that the parallelogram ABCD circumscribing a circle with center O is a rhombus.
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