CBSE Class 10 Mathematics Basic 2025: question 34(b)

Hints

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Step-by-step solution

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1. Figure: $\Delta ABC$ with P on AB and Q on AC, PQ ∥ BC. Median AD meets BC at D (BD = DC) and cuts PQ at R.

   ½ mark

2. Since PQ ∥ BC, PR ∥ BD. In $\Delta ARP$ and $\Delta ADB$: $\angle PAR=\angle BAD$ [common], $\angle APR=\angle ABD$ [corresponding angles]. So $\Delta ARP\sim \Delta ADB$ [AA similarity].

   1 mark

3. Hence $\frac{AR}{AD}=\frac{PR}{BD}$ ...(i)

   ½ mark

4. Similarly, in $\Delta ARQ$ and $\Delta ADC$: $\angle QAR=\angle CAD$ [common], $\angle AQR=\angle ACD$ [corresponding angles], so $\Delta ARQ\sim \Delta ADC$ [AA]. This gives $\frac{AR}{AD}=\frac{RQ}{DC}$ ...(ii)

   1 mark

5. From (i) and (ii): $\frac{PR}{BD}=\frac{RQ}{DC}$

   ½ mark

6. AD is a median, so $BD=DC$. Therefore $PR=RQ$.

   1 mark

7. Hence R is the mid-point of PQ, i.e. the median AD bisects PQ.

   ½ mark

Answer: $PR=RQ$, so the median AD bisects PQ.

Revise the chapter

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