Question

State true or false:

A parallelogram $ABCD$ has $P$ the mid-point of $DC$ and $Q$ a point of $AC$ such that $CQ=\frac{1}{4}AC$. $PQ$ produced meets $BC$ at $R$.
Can it be concluded that.

$PR=\frac{1}{2}DB$ ?

• A. True
• B. False

Answer 1

The correct option is A True

Given: ABCD is a parallelogram. P is the mid point of CD.
Q is the point on AC such that $CQ=\frac{1}{4}AC$
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, $OC=\frac{1}{2}AC$
$OQ=OC-CQ$
$OQ=\frac{1}{2}AC-\frac{1}{4}AC$
$OQ=\frac{1}{4}AC$
$OQ=CQ$
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,
$PQ\parallel OD$ or $PR\parallel OD$
Now, In $△BCD$,
P is the mid point of CD and $PR\parallel BD$,
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 $PQ\parallel OD$
Thus, By mid point theorem, $PQ=\frac{1}{2}OD$
Similarly In $△BOC$,
$QR=\frac{1}{2}OB$
Add both the equations,
$PQ+QR=\frac{1}{2}(OD+OB)$
$PR=\frac{1}{2}\left(BD\right)$