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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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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.