CBSE Class 10 Mathematics Standard 2026: question 34(A)

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Step-by-step solution

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1. Statement (Basic Proportionality Theorem): If a line is drawn parallel to one side of a triangle so that it meets the other two sides at distinct points, then it divides those two sides in the same ratio.

   1 mark

2. Figure: Draw $△ABC$. Draw a line parallel to $BC$ that meets $AB$ at $D$ and $AC$ at $E$.

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

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

   Construction: Join $BE$ and $CD$. Draw $DM\perp AC$ (meeting $AC$ at $M$) and $EN\perp AB$ (meeting $AB$ at $N$).

   2 marks

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

   Take $EN$ as the height:
   $$\text{ar}(ADE)=\frac{1}{2}\times AD\times EN,\text{ar}(BDE)=\frac{1}{2}\times DB\times EN$$
   $$\therefore \frac{\text{ar}(ADE)}{\text{ar}(BDE)}=\frac{AD}{DB}\text{...(1)}$$

   Take $DM$ as the height:
   $$\text{ar}(ADE)=\frac{1}{2}\times AE\times DM,\text{ar}(DEC)=\frac{1}{2}\times EC\times DM$$
   $$\therefore \frac{\text{ar}(ADE)}{\text{ar}(DEC)}=\frac{AE}{EC}\text{...(2)}$$

   1 mark

4. $△BDE$ and $△DEC$ stand on the same base $DE$ and lie between the same parallels $DE$ and $BC$. Hence
   $$\text{ar}(BDE)=\text{ar}(DEC)\text{...(3)}$$

   From (1), (2) and (3), the left-hand sides of (1) and (2) are equal. Therefore
   $$\frac{AD}{DB}=\frac{AE}{EC}$$

   Hence proved.

   1 mark

Answer: Statement of BPT with proof that $\frac{AD}{DB}=\frac{AE}{EC}$

Revise the chapter

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