Question

BP = 2AC
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, $△APB\sim △ECB$
$\frac{AB}{EB}=\frac{BP}{BC}$
$\frac{2EB}{EB}=\frac{PB}{BC}$ (E is the mid point of AB)
$BP=2BC$
$BP=2AD$ (Since, BC = AD)