Question

O is mid-point of AP
If the above statement is true then mention answer as 1, else mention 0 if false

Answer 1

In $△APB$ and $△ECB$,
$\angle ABP=\angle EBC$ (Common angle)
$\angle PAB=\angle CEB$ (Corresponding angle of parallel lines)
$\angle BPA=\angle ECB$ (third angle)
Thus, $△APB\sim △ECB$ by AAA
$\frac{AB}{EB}=\frac{BP}{BC}$
$\frac{2EB}{EB}=\frac{PB}{BC}$ (E is the mid point of AB)
$BP=2BC$
$BC+CP=2BC$
$CP=BC$ or $PC=AD$ (Since BC = AD)
Now, In $△$s AOD and COP,
$\angle AOD=\angle COP$ (Vertically opposite angles)
$AD=PC$ (Proved above)
$\angle ADO=\angle PCO$ (Alternate angles for parallel lines BC and AD)
Thus, $△AOD\cong △POC$ (SAS rule)
Hence, $OA=OP$
or O is the mid point of AP.