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

Given:

\( \mathrm{O} \) is the point of intersection of the diagonals \( \mathrm{AC} \) and \( \mathrm{BD} \) of a trapezium \( \mathrm{ABCD} \) with \( \mathrm{AB} \| \mathrm{DC} \). Through \( \mathrm{O} \), a line segment \( \mathrm{PQ} \) is drawn parallel to \( \mathrm{AB} \) meeting \( \mathrm{AD} \) in \( \mathrm{P} \) and \( \mathrm{BC} \) in \( \mathrm{Q} \).

To do:

We have to prove that \( \mathrm{PO}=\mathrm{QO} \).

Solution:

In $\triangle A B D$ and $\triangle P O D, P O \| A B$

$\angle D =\angle D$       (Common angle)

$\angle A B D =\angle P O D$         (Corresponding angles)

Therefore, by AA similarity,

$\triangle A B D  \sim \triangle P O D$

This implies,

$\frac{O P}{A B}=\frac{P D}{A D}$........(i)

In $\triangle A B C$ and $\triangle O Q C, O Q \| A B$

$\angle C =\angle C$

$\angle B A C =\angle Q O C$            (Corresponding angles)

Therefore, by AA similarity,

$\triangle A B C  \sim \triangle O Q C$

This implies,

$\frac{O Q}{A B} =\frac{Q C}{B C}$..........(ii)

In $\triangle A D C, O P \| D C$

By basic proportionality theorem,

$\frac{A P}{P D}=\frac{O A}{O C}$........(iii)

In $\triangle A B C, O Q \| A B$

By basic proportionality theorem,

$\frac{B Q}{Q C}=\frac{O A}{O C}$..........(iv)

From (iii) and (iv), we get,
$\frac{A P}{P D}=\frac{B Q}{Q C}$

Adding 1 on both sides, we get,

$\frac{A P}{P D}+1 =\frac{B Q}{Q C}+1$

$\frac{A P+P D}{P D} =\frac{B Q+Q C}{Q C}$

$\frac{A D}{P D} =\frac{B C}{Q C}$

$\frac{P D}{A D}=\frac{Q C}{B C}$

$\frac{O P}{A B} =\frac{O Q}{B C}$          [From (i) and (ii)]

$\frac{O P}{A B} =\frac{O Q}{A B}$        [From (ii)]

$O P =O Q$ 

Hence proved.