$ABCD$ is a parallelogram, $E$ and $F$ are the mid points $AB$ and $CD$ respectively. $GFI$ is any line intersecting $AD, EF$ and $BC$ at $Q, P$ and $H$ respectively. Prove that $GP = PH$.
Given:
$ABCD$ is a parallelogram, $E$ and $F$ are the mid points $AB$ and $CD$ respectively. $GFI$ is any line intersecting $AD, EF$ and $BC$ at $Q, P$ and $H$ respectively.
To do:
We have to prove that $GP = PH$.
Solution:
![](/assets/questions/media/153848-53461-1633104598.png)
$E$ and $F$ are the mid-points of $AB$ and $CD$.
This implies,
$AE=E B=\frac{1}{2} A B$
$\mathrm{CF}=\mathrm{FD}=\frac{1}{2} \mathrm{CD}$
$\mathrm{AB}=\mathrm{CD}$ (Opposite sides of a parallelogram are equal)
$\frac{1}{2} \mathrm{AB}=\frac{1}{2} \mathrm{CD}$
$\mathrm{EB}=\mathrm{CF}$
$\mathrm{EB} \parallel \mathrm{CF}$
This implies,
$\mathrm{BEFC}$ is a parallelogram.
$\mathrm{BC} \| \mathrm{EF}$
$\mathrm{BE}=\mathrm{PH}$.........(i)
Therefore,
$AEFD$ is a parallelogram.
$\mathrm{AE}=\mathrm{GP}$..........(ii)
$\mathrm{E}$ is the mid-point of $\mathrm{AB}$
This implies,
$\mathrm{AB}=\mathrm{BE}$............(iii)
From (i), (ii) and (iii), we get,
$\mathrm{GP}=\mathrm{PH}$
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