CBSE Class 10 Mathematics Basic 2025: question 33(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 in distinct points, then it divides those two sides in the same ratio.

   1 mark

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

   ½ mark

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

   ½ mark

4. Construction: Join $BE$ and $CD$. Draw $EN\perp AB$ (with $N$ on $AB$) and $DM\perp AC$ (with $M$ on $AC$).

   ½ mark

5. Figure: Draw $△ABC$ with $A$ at the top. Mark $D$ on $AB$ and $E$ on $AC$ with $DE\parallel BC$. Show the segments $BE$ and $CD$, and the perpendiculars $EN$ to $AB$ and $DM$ to $AC$.

   ½ mark

6. Proof: Take $EN$ as the height. Then $\text{ar}(△ADE)=\frac{1}{2}\cdot AD\cdot EN$ and $\text{ar}(△BDE)=\frac{1}{2}\cdot DB\cdot EN$.

   So $\frac{\text{ar}(ADE)}{\text{ar}(BDE)}=\frac{AD}{DB}$ ...(1)

   Now take $DM$ as the height. Then $\text{ar}(△ADE)=\frac{1}{2}\cdot AE\cdot DM$ and $\text{ar}(△DEC)=\frac{1}{2}\cdot EC\cdot DM$.

   So $\frac{\text{ar}(ADE)}{\text{ar}(DEC)}=\frac{AE}{EC}$ ...(2)

   1 mark

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

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

   1 mark

Answer: Statement of BPT and proof that $\frac{AD}{DB}=\frac{AE}{EC}$ when $DE\parallel BC$

Revise the chapter

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