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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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