CBSE Class 10 Mathematics Standard 2025: question 33

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Step-by-step solution

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1. Given: In $△ABC$, a line $DE\parallel BC$ meets AB at D and AC at E. To prove: $\frac{AD}{DB}=\frac{AE}{EC}$. Construction: Join BE and CD; draw $EN\perp AB$ and $DM\perp AC$. (Figure: triangle ABC with D on AB, E on AC, DE parallel to BC, segments BE and CD, perpendiculars EN and DM.)

   2 marks

2. Proof: $\text{ar}(ADE)=\frac{1}{2}\cdot AD\cdot EN$ and $\text{ar}(BDE)=\frac{1}{2}\cdot DB\cdot EN$, so $\frac{\text{ar}(ADE)}{\text{ar}(BDE)}=\frac{AD}{DB}$ ...(1). Also $\text{ar}(ADE)=\frac{1}{2}\cdot AE\cdot DM$ 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

3. $△BDE$ and $△DEC$ are on the same base DE and between the same parallels DE and BC, so $\text{ar}(BDE)=\text{ar}(DEC)$ ...(3). From (1), (2) and (3): $\frac{AD}{DB}=\frac{AE}{EC}$. Hence proved.

   1 mark

4. Converse: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

   1 mark

Answer: $\frac{AD}{DB}=\frac{AE}{EC}$ (proved). Converse: if a line divides two sides of a triangle in the same ratio, it is parallel to the third side.

Revise the chapter

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