Question

The adjoining figure shows a trapezium ABCD in which side AB is parallel to side DC. P and Q are mid - points of diagonals AC and BD respectively. CQ joined and produced meets AB at point R.
Which of the following options are correct?

• A. PQ = $\frac{1}{4}$(AB + DC)
• B. PQ = $\frac{1}{2}$(AB + DC)
• C. PQ = $\frac{1}{3}$(AB -DC)
• D. PQ is parallel to AB

Answer 1

The correct option is D

PQ is parallel to AB

By ASA, $\Delta CDQ\cong \Delta RBQ$
$\Rightarrow$ CQ = QR and DC = RB
(i) In $\Delta$ ACR, P is mid-point of AC and Q is mid-point of RC
$\Rightarrow$ PQ || AR $\Rightarrow$ PQ || AB
(ii) In $\Delta$ ACR, P is mid-point of AC and Q is mid-point of CR
$\Rightarrow PQ=\frac{1}{2}AR=\frac{1}{2}(AB-RB)$
$=\frac{1}{2}(AB-DC)[\therefore RB=DC]$.