[SQP] In 𝛥ABC, P and Q are points on AB & AC such that PQ is parallel

Question 26 (B) In 𝛥ABC, P and Q are points on AB and AC respectively such that PQ is parallel to BC. Prove that the median AD drawn from A on BC bisects PQ. We need to prove R is mid-point of PQ i.e. PR = RQ Since PQ ∥ BC And AB is transversal ∠ APR = ∠ ABD Similarly, PQ ∥ BC And AC is transversal ∠ AQR = ∠ ACD In Δ APR & Δ ABD ∠ PAR = ∠ BAD ∠ APR = ∠ ABD Δ APR ~ Δ ABD Since ratio of sides of similar triangle are equal 𝑨𝑷/𝑨𝑩=𝑷𝑹/𝑩𝑫 In Δ AQR & Δ ACD ∠ QAR = ∠ CAD ∠ AQR = ∠ ACD Δ AQR ~ Δ ACD Since ratio of sides of similar triangle are equal 𝑨𝑸/𝑨𝑪=𝑸𝑹/𝑪𝑫 Also, In Δ APQ & Δ ABC ∠ APQ = ∠ ABC ∠ AQP = ∠ ACB Δ APQ ~ Δ ABC Since ratio of sides of similar triangle are equal 𝑨𝑷/𝑨𝑩=𝑨𝑸/𝑨𝑪 And from (1) and (2) 𝑨𝑷/𝑨𝑩=𝑷𝑹/𝑩𝑫 & 𝑨𝑸/𝑨𝑪=𝑸𝑹/𝑪𝑫 From (1), (2) and (3) 𝑷𝑹/𝑩𝑫=𝑸𝑹/𝑪𝑫 Since BD = CD given 𝑃𝑅/𝐵𝐷=𝑄𝑅/𝐵𝐷 PR = QR Hence proved

Question 26 (B) - CBSE Class 10 Sample Paper for 2025 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

Question 26 (B)

In 𝛥ABC, P and Q are points on AB and AC respectively such that PQ is parallel to BC. Prove that the median AD drawn from A on BC bisects PQ.

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