O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB || DC. Through O, a line segment PQ is drawn parallel AB meeting AD in P and BC in Q. Prove that PO = QO.
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Given: In trapezium ABCD, AB || DCDiagonals BD and AC intersect at O and POQ || DC || AB
To Prove: PO = QO
Proof: In $\triangle\text{ABD},$
$\text{PO}\parallel\text{AB}$ [Given]
$\therefore\frac{\text{AP}}{\text{PD}}=\frac{\text{BO}}{\text{OD}}\ .....(\text{i})$
Similarly, in $\triangle\text{BDC},$
$\text{OQ}\parallel\text{DC}$
$\therefore\frac{\text{BO}}{\text{OD}}=\frac{\text{BQ}}{\text{QC}}\ .....(\text{ii})$
From (i) and (ii), we have
$\frac{\text{AP}}{\text{PD}}=\frac{\text{BQ}}{\text{QC}}$
$\Rightarrow\frac{\text{AP}}{\text{PD}}+1=\frac{\text{BQ}}{\text{QC}}+1$ [Adding 1 on both sides]
$\Rightarrow\frac{\text{AP+PD}}{\text{PD}}=\frac{\text{BQ+QC}}{\text{QC}}$
$\Rightarrow\frac{\text{AD}}{\text{PD}}=\frac{\text{BC}}{\text{QC}}\text{ or }\frac{\text{PD}}{\text{AD}}=\frac{\text{QC}}{\text{BC}}\ .....(\text{iii})$
In $\triangle\text{DOP}$ and $\triangle\text{DBA},$
$\text{AB}\parallel\text{PO}$ [Given]
$\therefore\angle\text{DPO}=\angle\text{DAB}$ [Corresponding angles]
$\therefore\angle\text{DOP}=\angle\text{DBA}$ [Corresponding angles]
$\therefore\triangle\text{DOP}\sim\triangle\text{DBA}$
$\Rightarrow\frac{\text{PO}}{\text{AB}}=\frac{\text{DP}}{\text{DA}}\ .....(\text{iv})$
Similarly, $\triangle\text{COQ}\sim\triangle\text{CAB}$ [By AA similarity criterion]
$\therefore\frac{\text{OQ}}{\text{AB}}=\frac{\text{QC}}{\text{BC}}\ .....(\text{v})$
From (iii), (iv) and (v), we have
$\frac{\text{PO}}{\text{AB}}=\frac{\text{OQ}}{\text{AB}} $
$\Rightarrow\text{PO}={\text{OQ}}$
Hence, proved.
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