Question

State true or false:
The following figure shows a trapezium $ABCD$ in which $AB\parallel DC$. $P$ is the mid-point of $AD$ and $PR\parallel AB$.
Hence,
$PR=\frac{1}{2}(AB+CD)$

• A. True
• B. False

Answer 1

The correct option is A True

In $△$, BDC and BQR,
$\angle B=\angle B$ (Common)
$\angle BDC=\angle BQR$ (Corresponding angles of parallel lines)
$\angle BCD=\angle BRQ$ (Corresponding angles of parallel lines)
thus, $△BDC\sim △BQR$
Thus, $\frac{BD}{QB}=\frac{DC}{QR}$
$2=\frac{DC}{QR}$ (Q is the mid point of BD)
$QR=\frac{1}{2}DC$
Similarly, $QP=\frac{1}{2}AB$
Hence, $QR+QP=\frac{1}{2}(AB+DC)$
$PR=\frac{1}{2}(AB+CD)$