Question

ABCD is a parallelogram, E and $F$ are the mid-points of $AB$ and $CD$ respectively. G His any line intersecting $AD,$ EF and $BC$ at $G,P$ and $H$ respectively then $GP=PH$ .

• A. True
• B. False

Answer 1

The correct option is A True

$ABCD$ is a parallelogram, $E$ and $F$ are mid points of $AB$ and $DC$ and a line $GH$ intersects $AD,EF$ and $BC$ at $G,P$ and $H$ respectively.

Join $GB$ and let it intersect $EF$ of $X.$

Now since $E$ and $F$ are mid points of $AB$ and $DC$

$\Rightarrow$ $AE=EB=\frac{1}{2}AB$

$\Rightarrow$ and $DF=FC=\frac{1}{2}DC$

but $AB=DC$

$\Rightarrow$ $AEFD$ and $BCFE$ are parallelogram

$\Rightarrow$ $AD\parallel EF\parallel BC$ ----- ( 1 )

Now In $△ABG$

$AE=EB$ and $EX\parallel AG$ [ From ( 1 ) ]

$\Rightarrow$ $X$ is the mid point of $GB$ [mid point theorem ]

and in $△GBH$

$\Rightarrow$ $GX=XB$ [ $\because$ $X$ is the mid point of $GB$ ]

$\Rightarrow$ and $XP\parallel BH$ [ From ( 1 ) ]

$\Rightarrow$ $P$ is the mid point of $GH$

$\Rightarrow$ $GP=PH$