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, point $P$ bisects,

• A. $BE$
• B. $AB$
• C. $BC$
• D. none of the above

Answer 1

Answer: (A)

$BE$

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
To prove: P is mid point of BE.
In $△APB$ and $△EPD$
$\angle APB=\angle EPD$ (Vertically opposite angles)
$\angle EDP=\angle PAB$ (Alternate angles)
$PA=PD$ (P is mid point of AD)
Thus, $△APB\cong △DPE$ (ASA rule)
Hence, $PE=PB$ (By cpct)
thus, P is mid point of BE