CBSE Class 10 Mathematics Standard 2025: question 33(a)

Hints

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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 to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

   1 mark

2. Given: $△ABC$ in which 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 DE parallel to BC, BE, CD, EN, DM drawn.

   2 marks

3. Proof: $ar(ADE)=\frac{1}{2}\cdot AD\cdot EN$, $ar(BDE)=\frac{1}{2}\cdot DB\cdot EN$, so $\frac{ar(ADE)}{ar(BDE)}=\frac{AD}{DB}$ ...(1). Similarly $ar(ADE)=\frac{1}{2}\cdot AE\cdot DM$, $ar(DEC)=\frac{1}{2}\cdot EC\cdot DM$, so $\frac{ar(ADE)}{ar(DEC)}=\frac{AE}{EC}$ ...(2)

   1 mark

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

   1 mark

Answer: Converse = Basic Proportionality Theorem: a line parallel to one side divides the other two sides in the same ratio (proved by area method)

Revise the chapter

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