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Question

Prove that the parallelogram circumscribing a circle is a rhombus.


Solution

Since ABCD is a parallelogram circumscribed in a circle

$$AB=CD........(1)$$

$$BC=AD........(2)$$

$$DR=DS$$ (Tangents on the circle from same point $$D$$)

$$ CR=CQ$$(Tangent on the circle from same point $$C$$)

$$ BP=BQ$$ (Tangent on the circle from same point $$ B$$ )

$$AP=AS$$ (Tangents on the circle from same point $$A$$)

Adding all these equations we get

$$DR+CR+BP+AP=DS+CQ+BQ+AS$$

$$(DR+CR)+(BP+AP)=(CQ+BQ)+(DS+AS)$$

$$CD+AB= AD+BC$$

Putting the value of equation $$1$$ and $$2$$ in the above equation we get

$$2AB=2BC$$

$$AB=BC...........(3)$$

From equation $$(1)$$, $$(2)$$ and $$(3)$$ we get

$$AB=BC=CD=DA$$

$$\therefore ABCD$$ is a Rhombus

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Mathematics
RS Agarwal
Standard X

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