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SONU

· started a discussion

· 1 Months ago

explian (Question: 184407.0 )

sir plz explain this

Question:
ABCD is a parallelogram. P and R are the mid points of DC and BC respectively. The line PR intersects the diagonal AC at Q. The length of CQ is equal to–
Options:
A) \(\cfrac{AC}{4}\).
B) \(\cfrac{BD}{3}\).
C) \(\cfrac{BD}{4}\).
D) \(\cfrac{AC}{3}\).
Solution:
Ans: (a) \(\cfrac{AC}{4}\).

Knowledge Expert

· commented

· 1 Months ago

Theorem: 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, then R is the midpoint of BC.
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)
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