$ \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.
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