$ABCD$ 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:

$E$ 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,

$\mathrm{GP}=\mathrm{PH}$

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

54 Views

Kickstart Your Career

Get certified by completing the course

Get Started