CBSE Class 10 Mathematics Standard 2024: question 34(A)

Hints

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Step-by-step solution

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1. Figure: Draw $△ABC$. Take a line parallel to $BC$ that meets $AB$ at $D$ and $AC$ at $E$. Join $BE$ and $CD$. Draw $EN\perp AB$ (N on AB) and $DM\perp AC$ (M on AC).

   ½ mark

2. Given: In $△ABC$, $DE\parallel BC$, with $DE$ meeting $AB$ at $D$ and $AC$ at $E$.

   ½ mark

3. To prove: $\frac{AD}{DB}=\frac{AE}{EC}$

   ½ mark

4. Construction: Join $BE$ and $CD$. Draw $EN\perp AB$ and $DM\perp AC$.

   ½ mark

5. Proof: Use area $=\frac{1}{2}\times$ base $\times$ height.

   In $△ADE$ and $△BDE$, the height from $E$ is $EN$ for both:

   $\frac{ar(ADE)}{ar(BDE)}=\frac{\frac{1}{2}\cdot AD\cdot EN}{\frac{1}{2}\cdot DB\cdot EN}=\frac{AD}{DB}$ ... (1)

   1 mark

6. In $△ADE$ and $△DEC$, the height from $D$ is $DM$ for both:

   $\frac{ar(ADE)}{ar(DEC)}=\frac{\frac{1}{2}\cdot AE\cdot DM}{\frac{1}{2}\cdot EC\cdot DM}=\frac{AE}{EC}$ ... (2)

   1 mark

7. $△BDE$ and $△DEC$ have the same base $DE$ and lie between the same parallels $DE$ and $BC$. So $ar(BDE)=ar(DEC)$ ... (3)

   From (1), (2) and (3): $\frac{AD}{DB}=\frac{AE}{EC}$. Hence proved.

   1 mark

Answer: Proved: if DE ∥ BC in △ABC, then AD/DB = AE/EC (Basic Proportionality Theorem).

Revise the chapter

• Chapter: Triangles (notes, formulas, where marks are lost and every past question)
• Topic: Basic Proportionality Theorem (Thales Theorem)