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Circles
RD Sharma
CBSE
Grade 10
Question:
State and Prove Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal. Use this theorem to prove:
If a circle touches all four sides of a quadrilateral $ABCD$, prove that $AB + CD = AD + BC$. Further, if $ABCD$ is a parallelogram, prove that it is a rhombus.
If a circle touches all four sides of a quadrilateral $ABCD$, prove that $AB + CD = AD + BC$. Further, if $ABCD$ is a parallelogram, prove that it is a rhombus.
Step-by-Step Solution
Key Concept: Part 1: Proof of Theorem 10.2 ($TP = TQ$) using RHS congruence of $\Delta OPT$ and $\Delta OQT$. Part 2: Add 4 tangent equations $\Rightarrow AB + CD = AD + BC$. Part 3: Since $AB = CD$ and $AD = BC$, $2AB = 2AD \Rightarrow AB = AD \Rightarrow ABCD$ is a rhombus.
Part 1: Complete Statement & Proof of Theorem 10.2 ($TP = TQ$ via RHS congruence). [2.0 Marks]
Part 2: Tangents from vertices: $AP = AS, BP = BQ, CR = CQ, DR = DS$. Add: $AB + CD = AD + BC$. Proved! [1.5 Marks]
Part 3: In parallelogram $ABCD$, $AB = CD$ and $AD = BC \Rightarrow 2AB = 2AD \Rightarrow AB = AD$. Since adjacent sides are equal, $ABCD$ is a rhombus. Proved! [1.5 Marks]
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🎯 Official CBSE Marking Scheme:
Proof of Theorem 10.2 ($TP = TQ$): 2.0 Marks
Proof of $AB + CD = AD + BC$: 1.5 Marks
Proof of circumscribed parallelogram is a rhombus: 1.5 Marks
Part 2: Tangents from vertices: $AP = AS, BP = BQ, CR = CQ, DR = DS$. Add: $AB + CD = AD + BC$. Proved! [1.5 Marks]
Part 3: In parallelogram $ABCD$, $AB = CD$ and $AD = BC \Rightarrow 2AB = 2AD \Rightarrow AB = AD$. Since adjacent sides are equal, $ABCD$ is a rhombus. Proved! [1.5 Marks]
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🎯 Official CBSE Marking Scheme:
Proof of Theorem 10.2 ($TP = TQ$): 2.0 Marks
Proof of $AB + CD = AD + BC$: 1.5 Marks
Proof of circumscribed parallelogram is a rhombus: 1.5 Marks
Correct Answer:
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