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Question
The side AC of a triangle ABC is produced to point E so that CE = 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 meet AC at point P and EF at point R respectively.
Prove that:
- 3DF = EF
- 4CR = AB
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Solution
Consider the figure :
Here D is the midpoint of BC and DP is parallel to AB, therefore P is the midpoint of AC and
PD = `[1]/[2]`AB
(i) Again from the triangle AEF, we have AF || PD || CR and AP = `[1]/[3]`AE
Therefore, DF = `[1]/[3]` EF or we can say that 3DF = EF.
Hence, it is shown.
(ii) From the triangle PED, we have PD || CR and C is the midpoint of PE, therefore, CR = `[1]/[2]`PD
Now,
PD = `1/2` AB
`1/2"PD" = 1/4`AB
CR = `1/4`AB
4CR = AB
Hence, it is shown.
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