Question

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

Answer 1

$ABCD$ is a `trapezium where $AB\parallel DC$ and diagonals $AC$ and $BD$ intersect at $O.$

Let us draw a line $EF\parallel AB\parallel DC$ passing through point $O$

Now, in $△ADC,$

$EO\parallel DC$ [ Since, $EF\parallel DC$ ]

Line draw parallel to one side of triangle, intersects the other two sides im distinct points, then it divides the other $2$ side in same ratio.

So, $\frac{AE}{DE}=\frac{AO}{CO}$ ---- ( 1 )

Similarly, In $△DBA,$

$\Rightarrow$ $EO\parallel AB$ [ Since, $EF\parallel AB$ ]

Line draw parallel to one side of triangle, intersects the other two sides im distinct points, then it divides the other $2$ side in same ratio.

$\therefore$ $\frac{AE}{DE}=\frac{BO}{DO}$ ----- ( 2 )

From ( 1 ) and ( 2 ),

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

$\Rightarrow$ $\frac{AO}{BO}=\frac{CO}{DO}$ ----- Hence proved