Question

State true or false:

In parallelogram $ABCD$. $E$ is the mid-point of $AB$ and $AP$ is parallel to $EC$ which meets $DC$ at point $O$ and $BC$ produced at $P$. Hence

$BP=2AD$

• A. True
• B. False

Answer 1

The correct option is A True

In $△$s, APB and ECB,
$\angle ABP=\angle EBC$ (Common angle)
$\angle PAB=\angle CEB$ (Corresponding angles of parallel lines)
$\angle APB=\angle ECB$ (Third angle of the triangle)
Thus $△APB\sim △ECB$
Hence, $\frac{AB}{EB}=\frac{BP}{BC}$ (Corresponding sides of similar triangles)
$2=\frac{BP}{BC}$
$BP=2BC$
$BP=2AD$ (BC = AD)