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Triangles
RD Sharma
CBSE
Grade 10
Question:
Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio (Basic Proportionality Theorem). Use this theorem to prove the following:
In $\Delta ABC$, $DE \parallel BC$ and $CD \parallel EF$, where $E$ lies on $AC$ and $F$ lies on $AD$. Prove that $AD^2 = AF \cdot AB$.
In $\Delta ABC$, $DE \parallel BC$ and $CD \parallel EF$, where $E$ lies on $AC$ and $F$ lies on $AD$. Prove that $AD^2 = AF \cdot AB$.
Step-by-Step Solution
Key Concept: Part 1: Proof of BPT using area of triangles. Part 2: In $\Delta ABC$, $DE \parallel BC \Rightarrow \dfrac{AD}{AB} = \dfrac{AE}{AC}$. In $\Delta ADC$, $EF \parallel CD \Rightarrow \dfrac{AF}{AD} = \dfrac{AE}{AC}$. Equate: $\dfrac{AD}{AB} = \dfrac{AF}{AD} \Rightarrow AD^2 = AF \cdot AB$.
Part 1: Proof of BPT (Statement, construction of altitudes & area ratios $\text{Area}(ADE)/\text{Area}(BDE) = AD/DB$ and $\text{Area}(ADE)/\text{Area}(DEC) = AE/EC$). [2.5 Marks]
Part 2: In $\Delta ABC$, $DE \parallel BC \Rightarrow \dfrac{AD}{AB} = \dfrac{AE}{AC}$ (by BPT corollary). (1) [1.0 Mark]
In $\Delta ADC$, $FE \parallel CD \Rightarrow \dfrac{AF}{AD} = \dfrac{AE}{AC}$ (by BPT corollary). (2) [1.0 Mark]
From (1) and (2): $\dfrac{AD}{AB} = \dfrac{AF}{AD} \Rightarrow AD^2 = AF \cdot AB$. Proved! [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Complete proof of BPT: 2.5 Marks
Applying BPT in $\Delta ABC \Rightarrow AD/AB = AE/AC$: 1.0 Mark
Applying BPT in $\Delta ADC \Rightarrow AF/AD = AE/AC$: 1.0 Mark
Equating ratios to conclude $AD^2 = AF \cdot AB$: 0.5 Mark
Part 2: In $\Delta ABC$, $DE \parallel BC \Rightarrow \dfrac{AD}{AB} = \dfrac{AE}{AC}$ (by BPT corollary). (1) [1.0 Mark]
In $\Delta ADC$, $FE \parallel CD \Rightarrow \dfrac{AF}{AD} = \dfrac{AE}{AC}$ (by BPT corollary). (2) [1.0 Mark]
From (1) and (2): $\dfrac{AD}{AB} = \dfrac{AF}{AD} \Rightarrow AD^2 = AF \cdot AB$. Proved! [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Complete proof of BPT: 2.5 Marks
Applying BPT in $\Delta ABC \Rightarrow AD/AB = AE/AC$: 1.0 Mark
Applying BPT in $\Delta ADC \Rightarrow AF/AD = AE/AC$: 1.0 Mark
Equating ratios to conclude $AD^2 = AF \cdot AB$: 0.5 Mark
Correct Answer:
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