Prove that a parallelogram circumscribing a circle is a rhombus

0 votes

Best answer

**Solution**:

Let ABCD be the rhombus circumscribing the circle with centre O, such that AB, BC, CD and DA touch the circle at points P, Q, R and S respectively.

We know that the tangents drawn to a circle from an exterior point are equal in length.

∴ AP = AS………….(1)

BP = BQ……………(2)

CR = CQ …………...(3)

DR = DS……………(4).

Adding (1), (2), (3) and (4)

we get AP+BP+CR+DR = AS+BQ+CQ+DS

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

∴ AB+CD=AD+BC-----------(5)

Since AB=DC and AD=BC (opposite sides of parallelogram ABCD)

putting in (5) we get, 2AB=2AD or AB = AD.

∴ AB=BC=DC=AD

Since a parallelogram with equal adjacent sides is a rhombus, so ABCD is a rhombus.