Question

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

Answer 1

Given, $ABCD$ is a trapezium

Also, $AB\parallel DC$

Diagonals intersect at Point $O$,

Construct $OX$ to intersect $AD$ and $OX\parallel BA$

In $\Delta ADB$

As $AB\parallel XO$,

$\Rightarrow$ $\frac{DX}{XA}=\frac{DO}{BO}$ ...(i)

In $\Delta ADC$,

$CD\parallel AB$

$\Rightarrow$ $\frac{AX}{XD}=\frac{AD}{OC}$

$\Rightarrow$ $\frac{XD}{AX}=\frac{CO}{AO}$ ...(ii)

As LHS are equal equate RHS

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

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

Hence Proved.