Given: ABCD is a parallelogram. P is the mid point of CD. Q is the point on AC such that CQ=14AC PQ produced meets BC in R. Join BD, let BD intersect AC in O. O is the mid point of AC (Diagonals of parallelogram bisect each other) hence, OC=12AC OQ=OC−CQ OQ=12AC−14AC OQ=14AC OQ=CQ Therefore, Q is the mid point of OC. In △OCD, P is the mid point of CD and Q is the mid point of OC. BY mid point theorem, PQ∥OD or PR∥OD Now, In △BCD, P is the mid point of CD and PR∥BD, By converse of mid point theorem, R is the mid point of BC Now, In △AOD P is mid point of CD and Q is mid point of OC and PQ∥OD Thus, By mid point theorem, PQ=12OD Similarly In △BOC, QR=12OB Add both the equations, PQ+QR=12(OD+OB) PR=12(BD)
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Q1
Prove: PR=12DB
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Q2
A parallelogram ABCD has P the mid- point of DC and Q intersects AC such that CQ=14AC. PQ produced meets BC at R prove that : (a) R is the mid-point of BC (b) PR=12DB
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Q3
State true or false:
A parallelogram ABCD has P the mid-point of DC and Q a point of AC such that CQ=14AC. PQ produced meets BC at R. Can it be concluded that.
PR=12DB ?
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