Question

$ABCD$ is a parallelogram. $E$ is the mid-point of $AB$ and $F$ is the mid-point of $CD$. $GH$ is any line that intersects $AD,EF$ and $BC$ at $G,P$ and $H$ respectively. Prove that $GP=PH$.

Answer 1

$ABCD$ 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 midpoints of $AB$ and $DC$

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

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

$AB=DC$

$\Rightarrow AEFD$ and $BCFE$ are parallelogram

$\Rightarrow AD\parallel EF\parallel BC$....(i)

In $△ABG,AE=EB$ and $EX\parallel AG$

$\therefore X$ is the midpoint of $GB$

In $△GBH$,

$GX=XB$ [$X$ is the mid point of $GB$] and $XD\parallel BH$

$\therefore P$ is the midpoint of $GH\Rightarrow GP=PH$