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Question
Side AC of a ABC is produced to point E so that CE = `(1)/(2)"AC"`. D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meets AC at point P and EF at point R respectively. Prove that: 4CR = AB.
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Solution
In ΔDEP,
C and R are the mid-points of PE and DE respectively.
Also, DP || RC
∴ CR = `(1)/(2)"DP"`......(i)
In ΔABC,
D and P are the mid-points of BC andAC respectively.
Also, DP || AB
∴ DP = `(1)/(2)"AB"`......(ii)
Substituting the value of DP from (ii) and(i)
⇒ CR = `(1)/(2)(1/2 "AB")`
⇒ CR = `(1)/(4)"AB"`
∴ 4CR = AB.
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