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)

Suggest Corrections

111

Similar questions

ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that CQ = $\frac{1}{4}$ AC. If PQ produced meets BC at R, prove that R is the midpoint of BC.

Q. ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that

C Q = \frac{1}{4} A C .

If PQ produced meets BC at R, prove that R is the midpoint BC.

Q. In the above figure ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that CQ =

\frac{1}{4}

AC. If PQ produced meets BC at R, prove that R is the midpoint of BC.

In the figure, ABCD is a parallelogram in which P is the mid -point of DC and Q is a point on Ac such that $C Q = \frac{1}{4} A C$ If PQ produced meets BC at R, prove that R is a mid -point of BC.

Q. A parallelogram ABCD has P the mid- point of DC and Q intersects AC such that

C Q = \frac{1}{4} A C

. PQ produced meets BC at R prove that :
(a) R is the mid-point of BC
(b)

P R = \frac{1}{2} D B

Join BYJU'S Learning Program