ABCD is a trapezium in which $AB \| DC$ and its diagonals intersect each other at the point $O$. Show that $\frac{AO}{BO} = \frac{CO}{DO}$.

Given:

ABCD is a trapezium in which $AB \| DC$ and its diagonals intersect each other at the the point $O$. 

To do:

We have to show that $\frac{AO}{BO} = \frac{CO}{DO}$.

Solution:

Draw $EO \| DC$

In $\triangle ABD$,

$DC \| AB$

$EO \| DC$

 This implies,

$\frac{AE}{ED}=\frac{BO}{CO}$...........(i)

In $\triangle ADC$,

$EO \| DC$

This implies,

$\frac{AE}{ED}=\frac{AO}{CO}$.........(ii)

From (i) and (ii), we get,

$\frac{BO}{DO}=\frac{AO}{CO}$

$\frac{AO}{BO}=\frac{CO}{DO}$

Hence proved.

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Updated on: 10-Oct-2022

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