ABCD is parallelogramin which P is the midpoint of DC and Q is a point of AC such that CQ=1/4 AC. if PQ produced meets BC at R, prove that R is the midpoint of BC.

Asked by joseph | 26th Oct, 2012, 04:45: PM

Expert Answer:

 
Given: ABCD is a parallelogram. P is the mid point of CD.
Q is a point on AC such that CQ=(1/14)AC
PQ produced meet BC in R.
 
To prove : R is the mid point of BC
Construction : join BD in O.Let BD intersect AC in O.
 
Prove : O is the mid point of AC. {diagnols of parallelogram bisect each other }
 
Therefore OC = (1/2) AC
=> OQ = OC-CQ = (1/2)AC - (1/4)AC = (1/4)AC.
=> OQ = CQ 
 
therefore Q is the mid point of OC.
In triangle OCD,
P is the mid point of CD and Q is the mid point of OC,
therefore PQ is parallel to OD (Mid point theorem)
=> PR is parallel to BD
In traingle BCD,
P is the midpoint of CD and PR is parallel  to BD,
therefore R is the mid point of BC (Converse of mid point theorem)

Answered by  | 26th Oct, 2012, 07:24: PM

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