# $\mathrm{O}$ 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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Updated on: 10-Oct-2022

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