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

$O$ is mid-point of $AP$.

• 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 $△APB\sim △ECB$ (AAA rule)

Hence, $\frac{AB}{EB}=\frac{BP}{BC}$ (Corresponding sides of similar triangles)

$2=\frac{BP}{BC}$

$BP=2BC$
Now, in $△$s $OPC$ and $APB,$

$\angle OPC=\angle APB$ (Common angle)

$\angle POC=\angle PAB$ (Corresponding angles of parallel lines)

$\angle PCO=\angle PBA$ (Third angle of a triangle)

$△OPC\sim △APB$ (AAA rule)

hence, $\frac{PC}{BP}=\frac{OP}{AP}$ (Corresponding sides)

$\frac{1}{2}=\frac{OP}{AP}$

$OP=\frac{1}{2}AP$

hence, $O$ is the midpoint of $AP$.