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 RhombusMathematicsRS AgarwalStandard X

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