$O$ is the point of intersection of the diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB \parallel DC$. Through $O$,a line segment $PQ$ is drawn parallel to $AB$ meeting $AD$ in $P$ and $BC$ in $Q$. Prove that $PO = QO$.

(N/A) Given: $ABCD$ is a trapezium with $AB \parallel DC$. Diagonals $AC$ and $BD$ intersect at $O$. $PQ \parallel AB \parallel DC$.
To prove: $PO = QO$.
Proof: In $\triangle ADC$,$PO \parallel DC$ (since $PQ \parallel DC$).
By the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{AP}{PD} = \frac{AO}{OC}$ ........$(i)$
In $\triangle ABC$,$OQ \parallel AB$.
By $BPT$,we have:
$\frac{BQ}{QC} = \frac{AO}{OC}$ ........$(ii)$
From $(i)$ and $(ii)$,we get:
$\frac{AP}{PD} = \frac{BQ}{QC}$
Adding $1$ to both sides:
$\frac{AP}{PD} + 1 = \frac{BQ}{QC} + 1$
$\frac{AP + PD}{PD} = \frac{BQ + QC}{QC}$
$\frac{AD}{PD} = \frac{BC}{QC}$
$\Rightarrow \frac{PD}{AD} = \frac{QC}{BC}$ ........$(iii)$
Now,in $\triangle ABD$,$PO \parallel AB$. Therefore,$\triangle POD \sim \triangle ABD$ (by $AA$ similarity).
So,$\frac{PO}{AB} = \frac{PD}{AD}$ ........$(iv)$
In $\triangle ABC$,$OQ \parallel AB$. Therefore,$\triangle OQC \sim \triangle ABC$ (by $AA$ similarity).
So,$\frac{OQ}{AB} = \frac{QC}{BC}$ ........$(v)$
From $(iii)$,$(iv)$,and $(v)$,we get:
$\frac{PO}{AB} = \frac{OQ}{AB}$
$\Rightarrow PO = OQ$. Hence proved.