CBSE Class 10 Mathematics Standard 2023: question 32

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Step-by-step solution

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1. Given: In $△ABC$, a line parallel to $BC$ meets $AB$ at $D$ and $AC$ at $E$, so $DE\parallel BC$.

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

   1 mark

2. Figure and construction: Draw $△ABC$ with $DE\parallel BC$, where $D$ is on $AB$ and $E$ is on $AC$. Join $BE$ and $CD$. Draw $EN\perp AB$ (N on AB) and $DM\perp AC$ (M on AC).

   1 mark

3. Proof: Using area $=\frac{1}{2}\times$ base $\times$ height:

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

   Similarly:

   • $\mathrm{ar}(ADE)=\frac{1}{2}\cdot AE\cdot DM$ and $\mathrm{ar}(DEC)=\frac{1}{2}\cdot EC\cdot DM$
   • So $\frac{\mathrm{ar}(ADE)}{\mathrm{ar}(DEC)}=\frac{AE}{EC}$ ...(2)
   1 mark

4. $△BDE$ and $△DEC$ stand on the same base $DE$ and lie between the same parallels $DE$ and $BC$.

   So $\mathrm{ar}(BDE)=\mathrm{ar}(DEC)$ ...(3)

   1 mark

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

   Hence the other two sides are divided in the same ratio. Hence proved.

   1 mark

Answer: $\frac{AD}{DB}=\frac{AE}{EC}$ (Basic Proportionality Theorem proved)

Revise the chapter

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