CBSE Class 10 Mathematics Basic 2024: question 34(b)

Hints

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Step-by-step solution

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1. Figure: $△ABC$ with a line DE ∥ BC meeting AB at D and AC at E.
   Given: In $△ABC$, $DE\parallel BC$. To prove: $\frac{AD}{DB}=\frac{AE}{EC}$.

   ½ mark

2. Construction: Join BE and CD. Draw $DM\perp AC$ and $EN\perp AB$.

   1 mark

3. $\frac{ar(△ADE)}{ar(△BDE)}=\frac{\frac{1}{2}\times AD\times EN}{\frac{1}{2}\times DB\times EN}=\frac{AD}{DB}$ ...(i)

   1 mark

4. $\frac{ar(△ADE)}{ar(△CDE)}=\frac{\frac{1}{2}\times AE\times DM}{\frac{1}{2}\times EC\times DM}=\frac{AE}{EC}$ ...(ii)

   1 mark

5. $△BDE$ and $△CDE$ lie on the same base DE and between the same parallels DE and BC, so $ar(△BDE)=ar(△CDE)$ ...(iii)

   1 mark

6. From (i), (ii) and (iii): $\frac{AD}{DB}=\frac{AE}{EC}$. Hence proved.

   ½ 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)