Question

given below, ABCD is a parallelogram. E is mid-point
is a point on AC such that PC = 1 AC. EP produced meets BC at F. Pro
(1) F is mid-point of BC
(ii) 2EF = BD.
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Answer 1

Given: ABCD is a parallelogram. E 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)