Prove that the parallelogram circumscribing a circle is a rhombus
Answers
Step-by-step explanation:
Prove that the parallelogram circumscribing a circle is a rhombus.
According to Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal.
Therefore,
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
On re-grouping,
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.
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