Question

In parallelogram $ABCD$ , $E$ is the mid-point at $AB$ and $AP$ is parallel to $EC$ which meet $DC$ at point $O$ and $BC$ produced at $P$. Prove that
(I) $BP=2AD$
(I) $O$ is mid-point of $AP$

Answer 1

Given:- $ABCD$ is $||gm$

$E$ is mid point

$AP||EC$

To prove: $BP=2AD$

Proof:

In $\Delta EBC$ and $\Delta ABP$

$\angle B=\angle B$ (common)

$\angle PAB=\angle CEB$ (corresponding angle)

$\therefore$ By $AA$ congruence

$\Delta ABP\cong \Delta EBC$

So,

$\frac{AB}{EB}=\frac{BP}{BC}$ [properties of similar $\Delta$]

$\frac{2EB}{EB}=\frac{BP}{BC}$ [$\because E$ is midpoint]

$\frac{2}{1}=\frac{BP}{AD}$ $[BC=AD$ opposite side of || gm are equal]

$BP=2AD$ Hence, proved.

To prove: (ii) $O$ is midpoint of $AP$

proof: In $\Delta OPC$ and $\Delta APB$

$\angle P=\angle P$ common

$\angle POC=\angle PAB$ ($DC||AB$ corresponding angles)

By $\Delta A$ congruence

$\Delta OPC\cong \Delta APB$

So, $\frac{OC}{AB}=\frac{PC}{BP}$

$\frac{OC}{AB}=\frac{PC}{BP}$

$\frac{OC}{AB}=\frac{PC}{2AD}$ [proved in part (i)]

$\frac{OC}{AB}=\frac{PC}{2BC}$.......(i) $[\because AD=BC]$

As $BP=2AD$

SO, $BP=2BC$ $[\because AD=BC]$

So,

$\because C$ is the midpoint of $BP$

Hence, $BC=PC$

Putting in equation (1)

$\frac{OC}{AB}=\frac{BC}{2BC}$

$AB=2OC$

$DC=2OC$ ($\because AB=DC$ opposite sides of || gm)

$\therefore O$ is mid point of $CD$

Hence, proved.