Question

In the given figure, ABCD is a rhombus, P and Q are the mid-points of BO and DO respectively. Which of the following is not always correct?

• A. $AQ=PC$
• B. $AP=QC$
• C. $AQ=QC$
• D. $OQ=\frac{1}{2}OA$

Answer 1

The correct option is D $OQ=\frac{1}{2}OA$

The diagonals of a rhombus bisect each other at right angles.
$\therefore OB=OD$ …..(i)
Since P and Q are the mid-points of BO and DO respectively.
$\therefore OP=PB$ and $OQ=QD$
$\Rightarrow OP=PB=OQ=QD$ [From (i)] …..(ii)
In $\Delta AOQ$ and $\Delta COP$,
$AO=OC$ (Diagonals bisect each other)
$\angle AOQ=\angle COP$ (Vertically opposite angles)
$OP=OQ$ [From (ii)]
$\therefore \Delta AOQ\cong \Delta COP$ (SAS congruence rule)
$\Rightarrow AQ=PC$ (CPCT)
Similarly, $\Delta QOC\cong \Delta POA$ (SAS Congruence rule)
$\Rightarrow AP=QC$ (CPCT)
In $\Delta AOQ$ and $\Delta COQ$,
$AO=OC$ (Diagonals bisect each other)
$\angle AOQ=\angle COQ$ (Each $90º$)
$OQ=OQ$ (Common)
$\therefore \Delta AOQ\cong \Delta COQ$ (SAS congruence rule)
$\Rightarrow AQ=QC$ (CPCT)
Thus, $AQ=PC,AP=QC$ and $AQ=QC$ is correct.
Hence, the correct answer is option (d).