Question

In the adjoining figure, $ABCD$ is a ||gm in which $E$ and $F$ are the midpoints of $AB$ and $CD$ respectively. If $GH$ is a line segment that cuts $AD,EF$ and $BC$ at $G,P$and $H$ respectively, prove that $GP=PH$.

Answer 1

Solution: Given, $ABCD$ is a parallelogram, in which $E$ and $F$ are the midpoints of $AB$ and $CD$ respectively.
$\therefore AE=BE=\frac{AB}{2}$ and $CF=DF=\frac{1}{2}CD$
But, $AB=CD$
$\therefore \frac{1}{2}AB=\frac{1}{2}CD$
$\Rightarrow BE=CF$ and also $BE\parallel CF$ [Since, $AB\parallel CD$]
Therefore, $BEFC$ is a parallelogram.
$\Rightarrow BC\parallel EF$ and $BE=PH$.....(i)
Since $AD\parallel BC$ as $ABCD$ is a prallelogram.
$AD\parallel EF$
Thus, $AEFD$ is a parallelogram.
$AE=GP$....(ii)
Since $AE=BE$.....(iii)
From equation(ii) and (iii), we get $BE=GP$....(iv)
From equation(i) and (iv), we get $GP=PH$
Hence, proved.