Q) Prove that the parallelogram circumscribing a circle is a rhombus.

9th Class Maths – NCERT Important Questions

Ans:

Step 1: Let’s make a diagram for better understanding of the question:

Here ABCD is the cyclic parallelogram. It is circumscribing a circle with center O.

Step 2:

∵ tangents from an external point to a circle are always equal

∴ tangents from point A, AP = AS

∴ tangents from point B, BP = BQ

∴ tangents from point C, CR = CQ

∴ tangents from point D, DR = DS

Step 3: By adding all equations from step 2, we get:

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

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

∴ AB + CD = AD + BC

Step 4: ∵ ABCD is a parallelogram

∴ Opposite sides of a parallelogram are always equal.

∴ AB = CD and AD = BC

Step 5: By substituting equations of step 4 in equation of step 3, we get:

∵ AB + CD = AD + BC

∴ AB + AB = AD + AD

∴ 2 AB = 2 AD

∴ AB = AD

Step 6:

Summing up all relations from step 4 & 5, we get:

∴ AB = BC = CD = AD

Therefore, ABCD is a Rhombus.

Hence Proved!

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