Prove that the parallelogram circumscribing a circle is a rhombus.
Step-by-Step Solution
Key Concept: Use circumscribed quadrilateral property $AB + CD = AD + BC$. Since $ABCD$ is a parallelogram, $AB = CD$ and $AD = BC \Rightarrow 2 AB = 2 AD \Rightarrow AB = AD$.
Let $ABCD$ be a parallelogram circumscribing a circle. We know for any circumscribed quadrilateral, $AB + CD = AD + BC$ -- (1). [1.0 Mark]
Since $ABCD$ is a parallelogram, opposite sides are equal: $AB = CD$ and $AD = BC$. [1.0 Mark]
Substitute in (1): $AB + AB = AD + AD \Rightarrow 2 AB = 2 AD \Rightarrow AB = AD$. [0.5 Mark]
Since adjacent sides are equal ($AB = AD = CD = BC$), parallelogram $ABCD$ is a RHOMBUS. Proved! [0.5 Mark]
---
🎯 Official CBSE Marking Scheme:
Stating $AB + CD = AD + BC$: 1.0 Mark
Applying parallelogram opposite sides equal ($AB=CD, AD=BC$): 1.0 Mark
Deducing $AB = AD$ and concluding rhombus: 1.0 Mark
Correct Answer: