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Question
In ΔABC, P is the mid-point of BC. A line through P and parallel to CA meets AB at point Q, and a line through Q and parallel to BC meets median AP at point R. Prove that: BC = 4QR
Solution
In ΔABC,
Q and S are the mid-points of AB and AC respectively. Also QS is parallel to BC
Therefore, QS = `(1)/(2)"BC"` ......(i)
Now, AP is the median, hence it bisects BC and QS
Therefore
`(1)/(2)"QS"` = QR
⇒ QS = 2QR
Substituting in (i)
⇒ 2QR = `(1)/(2)"BC"`
⇒ BC = 4QR.
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