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

Made by

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.

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