Question

State true or false:

In trapezium $ABCD$, $AB$ is parallel to $DC$; $P$ and $Q$ are the mid-points of $AD$ and $BC$ respectively. $BP$ produced meets $CD$ produced at point $E$. Hence, $PQ$ is parallel to $AB$

• A. True
• B. False

Answer 1

The correct option is A True

Given: ABCD is a trapezium. $AB\parallel DC$. P and Q are mid points of AD and BC respectively.
BP produced meets CD at E

Construction: Join BD. Draw a parallel line from P which meets BD on M such that $PM\parallel AB$ and a parallel line from Q which meets BD on N such that $QN\parallel CD$
Now, In $△ADB$
P is mid point of AD and $PM\parallel AB$. Thus, M is mid point of BD.
In $△BDC$
Q is mid point of BC and $QN\parallel DC$. Thus, N is mid point of BD
Hence, M and N are same points. Thus, PM or QN is a straight line, PQ
and $PQ\parallel AB\parallel DC$