0

Question

ABCD is a parallelogram in which P is the midpoint of DC and Q is point on AC such that CQ=1/4 AC . If produced PQ meets at BC at R then prove that R is the midpoint of BC.

Open in App

Solution

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)

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)

83

View More

Join BYJU'S Learning Program

Join BYJU'S Learning Program